Answer:
A) q = 1.67 × 10^(-7) C
B) q = 1.67 × 10^(-10) C
Explanation:
We are given;
Potential; V = 30 KV = 30000 V
Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m
A) To find the charge of the sphere, we will use the formula;
V = kq/r
Where;
q is the charge
k is electric force constant = 9 × 10^(9) N.m²/C²
Thus;
q = Vr/k
q = (30000 × 0.05)/(9 × 10^(9))
q = 1.67 × 10^(-7) C
B) Now, potential energy here is a formula; U = qV
However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;
qV = ½mv²
q = mv²/2V
Where;
v is speed = 10 m/s
V = 30000 V
m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;
q = (0.1 × 10^(-6) × 10²)/(2 × 30000)
q = 1.67 × 10^(-10) C
You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s
What are stepdown transformers used for
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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A car is moving with a velocity of45m/sis brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before it comes to rest.
Explanation:
Given,
v = 45m/s
t = 5s
Therefore,
d = v × t
= 45 × 5
= 225m
Define relative density.
Relative density is the ratio of the density of a substance to the density of a given material.
Displacement of a body moving in circular motion is
Explanation:
Displacement of a body moving in circular motion is called uniform circular motion.
hope it is helpful to you
Answer:
A constantly moving object with consistent circular movement. However, for its change in direction, it is accelerating
Explanation:
Uniform circular motion in a circle at constant rate can be described as the motion of the object. When an object moves in a circle, it changes its direction constantly. The object moves tangently to the circle at all times. As the velocity vector direction is the same as the object motion direction, the velocity vector is tangent to the circle. This is shown in the animation on the right by a vector arrow.
An item is accelerating that moves in a circle. Objects that accelerate are subjects that change their speed – either the velocity (i.e. the vecteur magnitude) or the direction. An object with consistent circular movement moves at a constant speed. However, because of its change in direction, it is accelerating. The acceleration direction is inside. The animation on the right shows this through a vector arrow
For an object with only a uniform circular movement, the final motion is the net force. The The net force acting on this object is directed to the middle of the circle. The net force is an inner or centripetal force. Without such a deepest force, an object would continue in a straight direction, never deviating. Regrettably, with the inward net force, perpendicular to the vector, the object changes the direction and is accelerated internally.
The value of mass remains constant but weight changes place to place why
Explanation:
No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).
Explanation:
Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W
Answer:
a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W
b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W
Explanation:
a. Find the current (in A) and power (in W) for each when connected in series.
Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.
So, substituting the values of the variables into the equation, we have
R = R₁ + R₂
R = 30.0 Ω + 88.0 Ω
R = 118.0 Ω
Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω
So, I = V/R = 64.0V/118.0 Ω = 0.542 A
Since the resistors are in series, the same current flows through them
i. Current in 30.0 Ω
Current in 30.0 Ω is I = 0.542 A since the resistors are in series.
ii Power in the 30.0 Ω
The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω
So, P₁ = I²R₁
= (0.542 A)² × 30.0 Ω
= 0.293764 A² × 30.0 Ω
= 8.8129 W
≅ 8.813 W
iii. Current in 88.0 Ω
Current in 88.0 Ω is I = 0.542 A since the resistors are in series.
iv. Power in the 88.0 Ω
The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω
So, P₂ = I²R₂
= (0.542 A)² × 88.0 Ω
= 0.293764 A² × 88.0 Ω
= 25.8512 W
≅ 25.85 W
(b) Repeat when the resistances are in parallel.
Since the resistors are connected in parallel, the same voltage is applied across them.
i. Current in 30.0 Ω
Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω
So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A
ii Power in the 30.0 Ω
The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω
So, P₁ = V²/R₁
P₁ = (64.0 V)²/30.0 Ω
P₁ = 4096 V²/30.0 Ω
P₁ = 136.53 W
iii. Current in 88.0 Ω
Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω
So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A
iv. Power in the 88.0 Ω
The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω
So, P₂ = V²/R₂
P₂ = (64.0 V)²/88.0 Ω
P₂ = 4096 V²/88.0 Ω
P₂ = 46.55 W
measurement is essential in our life.justify the statement.
Answer:
Measurements allow people to find their way to new places. Measurements such as miles or kilometers are used by GPS systems to give directions. Time measurements help to create schedules so tasks get done on time. Measurements are used in food as well. Ingredients in recipes have to be measured to make the dish correctly. Serving sizes are a measurement that keep people healthy by showing how much of each food you should eat.
~~~NEED HELP ASAP~~~
Please solve each section and show all work for each section.
Explanation:
Forces on Block A:
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as
[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1), we get
[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]
where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]
Forces on Block B:
Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as
[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]
From (5), we can solve for N as
[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]
Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by
[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]
Substituting (7) into (4) we get
[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]
Collecting similar terms together, we get
[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]
or
[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]
Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]
When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is 1.0 megohm (106 Ω). How much current passes through the insulation of the motor if the test voltage is 1000 V?
Answer:
0.001 A
Explanation:
Applying,
V = IR.............. Equation 1
Where V = Voltage of the motor, I = current, R = resistance
make I the subject of the equation
I = V/R.............. Equation 2
From the question,
Given: V = 1000 V, R = 1 MΩ = 10⁶ Ω
Substitute these values into equation 2
I = 1000/10⁶
I = 10⁻³ A
I = 0.001 A
Determine the absolute pressure on the bottom of a swimming pool 27.0 m by 8.9 m whose uniform depth is 1.8 m . Express your answer using two significant figures.
Answer:
[tex]P=17658Pa[/tex]
Explanation:
From the question we are told that:
Dimension
[tex]L*B=27.0*8.9[/tex]
Depth [tex]d=1.8m[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]V=L*B*D[/tex]
[tex]V=27.0*8.9*1.8[/tex]
[tex]V=432.54m^3[/tex]
Therefore
Force at the bottom of the Pool
[tex]F=\rho Vg[/tex]
Where
[tex]\rho \ density\ of \ water(1000kg/m^3)[/tex]
[tex]F=1000*432.54m^3*9.81[/tex]
[tex]F=4.2*10^{6}N[/tex]
Generally the equation for Pressure at the bottom is mathematically given by
[tex]P=\frac{Forece }{Area}[/tex]
[tex]P=\frac{4.2*10^{6}N}{27.0*8.9}[/tex]
[tex]P=17658Pa[/tex]
To calculate the gravitational potential energy of a statue on a 10-meters-tall platform, you would have to know the statue's ______________
Answer:
mass
Explanation:
A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?
Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
If an object of a constant mass experiences a constant net force, it will have a constant what?
Explanation:
hope it helps !!!!!!!!!!!!!
If an object of a constant mass experiences a constant net force, it will have a constant acceleration.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
According to Newton's second law of motion:
Applied force = mass × acceleration.
Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.
Learn more about force here:
https://brainly.com/question/26115859
#SPJ6
two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations
Answer:
The height is the same
Explanation:
Because they were at the same height but they fell at different velocities
The energy truck travelling at 10 km/h has kinetic energy. How much kinetic energy does it have when it is loaded so its mass is twice and its speed is increased to twice?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] is
[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]
When its mass and velocity are doubled, its new kinetic energy KE is
[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]
[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]
Therefore the kinetic energy will increase by a factor of 8.
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.
Answer:
KE = 2800 J
Explanation:
Usually a velocity is expressed as m/s. Then the energy units are joules.
[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]
v = 30 m / sec
KE = 1/2 * 4 * (30)^2
KE =2800 kg m^2/sec^2
KE = 2800 Joules
An object is moving from north to south what is the direction of the force of friction of the object
Answer:
North
Explanation:
Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.
Answer:
South To North
Explanation:
Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north
In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
You work in the special effects department of a movie studio. You are currently working on a superhero movie where the hero is very
strong and cannot be hurt by normal weapons such as a sword. In the next scene, the villain
is going to hit the hero with a lead pipe.
The hero's costume is mostly made of foam and carbon so that is it very light and easy to move around in but will crumble if hit with
anything hard. For this scene, a section of the
costume needs to be replaced with a different material that will only dent, not crumble,
when hit with the pipe. You are on a deadline and look around the props
department for ideas, what could you use?
A. You find sheets of copper (Cu) used in an induced rescue factory scene
B. You find a barrel of phosphorus (P) used as a component in explosive powder
C. You find a tank of helium (He) used for balloons
D. You find large desk of Chlorine (Cl) used to kill bacteria and pools
Answer:
A. You find sheets of copper (Cu) used in an induced rescue factory scene
Explanation:
I would use a sheets of copper (Cu) used in an induced rescue factory scene because, copper is a metal and the only material out of the other options that would only dent and not crumble when hit by the lead pipe.
The other options in B - D are non-metals.
Phosphorus is a non-metal and is used as an explosive and thus cannot be used.
Helium is a gas used in filling balloons and thus cannot be used.
Chlorine is used in killing bacteria and thus cannot be used.
So, copper is the only option available since it is a metal and can thus accommodate a large force.
So, option A is the answer.
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
Typhoon signal number 2 is raised. What is the speed of the expected typhoon?
the simple answer is from 61kmph to 120kmph
Explanation:
no explanation is needed
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
Answer:
Heat is flowing into the metal.
Explanation:
From the question given above, the following data were obtained:
Mass (M) of iron = 150 g
Initial temperature (T₁) = 25.0°C
Final temperature (T₂) = 73.3°C
Direction of heat flow =?
Next, we shall determine the change in the temperature of iron. This can be obtained as follow:
Initial temperature (T₁) = 25.0 °C
Final temperature (T₂) = 73.3 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 73.3 – 25
ΔT = 48.3 °C
Next, we shall determine the heat transfered. This can be obtained as follow:
Mass (M) of iron = 150 g
Change in temperature (ΔT) = 48.3 °C
Specific heat capacity (C) of iron = 0.450 J/gºC
Heat (Q) transfered =?
Q = MCΔT
Q = 150 × 0.450 × 48.3
Q = 3260.25 J
Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.
A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.645. What is the minimum stopping time for the truck in order to prevent the crate from sliding?
By Newton's second law,
• the net force acting vertically on the crate is 0, and
∑ F = n - mg = 0 ==> n = mg = 1470 N
where n is the magnitude of the normal force; and
• the net force acting in the horizontal direction on the crate is also 0, with
∑ F = f - b = 0 ==> b = f = µn = 0.645 (1470 N) = 948.15 N
where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.
Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.
Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :
f = ma ==> a = (948.15 N) / (150.0 kg) = 6.321 m/s²
With this acceleration, the truck will come to a stop after time t such that
0 = 50.0 km/h - (6.321 m/s²) t ==> t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s
and this is the smallest stopping time possible.
