Answer:
(a) The acceleration is 2.35 m/s^2.
(b) The time is 5.53 s.
Explanation:
coefficient of friction = 0.24
(a) The acceleration of the car is
[tex]a =\mu g\\\\a = 0.24 \times9.8\\\\a = 2.35 m/s^2[/tex]
(b) initial velocity, u =0, final velocity, v = 13 m/s
Let the time is t.
Use first equation of motion
v = u + a t
13 = 0 + 2.35 x t
t = 5.53 seconds
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
(a) v₁ = 51.96 km/h
(b) v₁ = 178 km/h
Explanation:
(a)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Cadillac = 2650 kg
v₂ = speed of Cadillac = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]
v₁ = 51.96 km/h
(b)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Truck = 9080 kg
v₂ = speed of Truck = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]
v₁ = 178 km/h
A 25N[L] force acts on a 1kg object that is already travelling at 10m/s[R]. The force acts for t=2s. What is its final speed? *
A) 25m/s
B) 25m/s/s
C) 40m/s[R]
D) 40m/s[L]
show your full work please
Answer:
C) 40m/s [R]
Explanation:
Given the following data;
Force = 25 N [L]
Mass = 1 kg
Initial velocity = 0 m/s
Velocity = 10 m/s [R]
Time = 2 seconds
To find the final velocity;
First of all, we would determine the acceleration of the object.
Force = mass * acceleration
25 = 1 * acceleration
Acceleration = 25/1
Acceleration = 25 m/s²
Next, we would determine the final velocity by using the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the formula, we have;
[tex] V = 0 + 25*2[/tex]
[tex] V = 0 + 50 [/tex]
V = 50 m/s
Final velocity = 50 - 10
Final velocity = 40 m/s [R]
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination
Answer:
The right solution is "4.8° east of north".
Explanation:
Given:
Distance,
= 500 km
Speed,
[tex]\vec{v}=120 \ m/s[/tex]
Wind (towards west),
[tex]v_0=10 \ m/s[/tex]
According to the question, we get
The angle will be:
⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]
[tex]=Cos^{-1}(\frac{10}{120} )[/tex]
[tex]=85.21[/tex] (north of east)
hence,
The direction must be:
⇒ [tex]\Theta'=90-85.21[/tex]
[tex]=4.79^{\circ}[/tex]
or,
[tex]=4.8^{\circ}[/tex] (east of north)
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F _ { x } = - [ 20.0 N + ( 3.0 N / m ) x ]F x =−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
[tex]W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J[/tex]
A 2.2 kg, 20-cm-diameter turntable rotates at 80 rpm on frictionless bearings. Two 600 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?
Answer:
[tex]w_2=38.3rpm[/tex]
Explanation:
From the question we are told that:
Mass of turntable [tex]M=2.2kg[/tex]
Diameter of turntable [tex]d=20cm=>0.2m[/tex]
Angular Velocity [tex]\omega =80rpm[/tex]
Mass of Blocks [tex]M_b=600g=>0.6kg[/tex]
Generally the equation for inertia is mathematically given by
Initial scenario at \omega=80rpm
[tex]I_1=\frac{1}{2}mR^2[/tex]
[tex]I_1=\frac{1}{2}*2.2*0.1^2[/tex]
[tex]I_1=0.11kgm^2[/tex]
Final scenario
[tex]I_2=I_1+2mR^2[/tex]
[tex]I_2=0.011+(2*0.6*0.12)[/tex]
[tex]I_2=0.023[/tex]
Generally the equation for The relationship between Angular velocity and inertia is mathematically given by
[tex]I_1w_1=I_2w_2[/tex]
[tex]w_2=\frac{I_1 \omega}{I_2}[/tex]
[tex]w_2=\frac{0.011*80}{0.023}[/tex]
[tex]w_2=38.3rpm[/tex]
The spring in the figure has a spring constant of 1400 N/m . It is compressed 17.0 cm , then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?
Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is
W = 1/2 kx ²
where k is the spring constant and x is the compression of the spring. So
W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J
This is equal to the block's change in kinetic energy ∆K,
W = ∆K
and since it starts from rest, the initial K is zero, leaving us with
W = 1/2 mv ²
where m is the mass of the block and v is its speed, so that
20.23 J = 1/2 (0.200 kg) v ²
==> v ≈ 14.2 m/s
The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.
First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.
Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:
• the net force acting on the block in the direction perpendicular to the incline is
∑ F = n - mg cos(45°) = 0
where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;
• the net force acting on the block parallel to the surface is
∑ F = -f - mg sin(45°) = ma
where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.
Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then
W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J
Use the work-energy theorem again to find the block's new speed v at the top of the incline:
W = ∆K
==> -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²
==> v ≈ 12.4 m/s
And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to
x = (12.4 m/s) cos(45°) t
Its height y from the 2.0 m-high surface at time t is given by
y = (12.4 m/s) sin(45°) t - 1/2 gt ²
The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.
The block sail through the air at the distance of "15.8 m"
Given:
Spring constant,
1400 N/mMass,
200 gBlock's coefficient,
0.210By using Work energy theorem, we get
→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]
By substituting the values, we get
→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]
here,
[tex]V_f = 12.44 \ m/s[/tex]
→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]
[tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]
[tex]= 15.8 \ m[/tex]
Thus the answer above is right.
Learn more:
https://brainly.com/question/13884105
Give a quantitative definition of being in contact.
Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.
At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:
A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.
Answer:
both swings at the same place
Explanation:
because there mother is giving same amount of force to both.
The number of current paths in a series circuit is:
a. one
b. two
C. three
d. four
Answer:
One
Explanation:
In series combination, the circuit follows one path whereas in parallel it follows two or more than two path
A temperature of 200 degrees Fahrenheit is equivalent to approximately A.93.3 degrees Celsius B. 232 degrees Celsius C. 37.8 degrees Celsius D. 840 degrees Celsius
Answer:
you can use G.oogle for this question.
Answer:
93.3 degrees Celsius.
Explanation:
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
a. V1 = V2
b. V1 > V2
c. V1 < V2
Answer:
We know that the change in electric potential energy is defined as:
q*ΔV = ΔP
So, the change in the electric potential energy is the charge times the change in the electric potential.
For the case of an electron gas, we have:
q = -e
where -e is the charge of an electron (remember that is negative).
So, if the electron gains electric potential then:
ΔP > 0
this means that the final potential energy is larger than the initial one, then we have:
-e*ΔV > 0
This means that ΔV must be negative.
V₂ = electric potential at point 2, so it is the final electric potential
V₁ = electric potential at point 1, so it is the final electric potential
Then we should get:
ΔV = V₂ - V₁ < 0.
This means that:
V₂ < V₁
The correct option is b.
Which of these statements is true?
a) Neither (a) nor (b) is true.
b) Both (a) and (b) are true.
c) Electric current is stored in the battery of a circuit.
d) Electric current is the flow of electrons in a circuit.
Answer:
d electric current is the flow of electrons in a circuit
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
here did you walk? What did you find most enjoyable while walking: listening to music, listening to an audio book, or nothing? How did your body react to this introductory amount of exercise? Was it more exercise or less exercise than you are used to? If you did not walk, what other type of physical movement did you do?
Which of the following refers to a force acting toward the center of a circular
motion?
A. Centrifugal force
B. Circular force
C. Central force
D. Centripetal force
Answer:
D. Centripetal Force :)
Verify that your equation has the masses and the velocities before and after the collision. Solve equation for the initial velocity of the projectile, Vo. As the bob swings upward from h1 to a max of h2, what is happening to the kinetic energy of the system?
Answer:
Decrease occur in kinetic energy.
Explanation:
The kinetic energy decreases when the the Bob swings reaches to the maximum height because the motion of the bob slower down. At maximum height, the kinetic energy decreases whereas the value of potential energy is the highest. The main reason of higher potential energy is that it depends on height of an object while on the other hand, kinetic energy depends on the motion of an object so that's why the value of kinetic energy decreases and potential energy increases at maximum height of the bob.
A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
a) how fast is book traveling the instant before it impacts the ground below the cliff?
b) how far above the bottom of the cliff is the object moving at 12 m/s?
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
[tex]v^2=v_0^2+2a[/tex]Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
[tex]v^2=0^2+2(-9.8)(-120)[/tex] so
[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
[tex]12^2=0^2+2(-9.8)[/tex]Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.
Leakage of liquid petroleum gas can be detected from a distance . name the process
Answer:
ultrasonic
Explanation:
they detect the acoustic emission created when a pressured gas expands in a low pressure area through a small orifice(the leak).
6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if its surface normal isin the XY plane and along a line that isinclined at 60 degrees to the positive Y axisand 30 degrees to the positive X axis
Answer:
Flux is 21 Nm^2/C.
Explanation:
Electric field, E = 6 N/C along X axis
Electric filed vector, E = 6 i N/C
Area, A = 4 square meter
Area vector
[tex]\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\[/tex]
The flux is given by
[tex]\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C[/tex]
Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.
Answer:
[tex]F = 14.2 N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15kg[/tex]
Radius [tex]r=0.6[/tex]
Angular Velocity [tex]\omega=2rev/s[/tex]
[tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]
Generally the equation for Force applied is mathematically given by
[tex]F =mrw2[/tex]
[tex]F=0.15*0.6* (4*x3.14^)2[/tex]
[tex]F = 14.2 N[/tex]
On a day when the speed of sound is 345 m/s, a 440 Hz tuning fork causes a tube closed at one end to vibrate in the second harmonic. How long is the tube?
Answer:
Length = 3.136 meters
Explanation:
Given the following data;
Speed = 345 m/s
Frequency = 440 Hz
To find how long is the tube;
First of all, we would determine the wavelength;
Wavelength = speed/frequency
Wavelength = 345/440
Wavelength = 0.784 m
Next, we would determine how long is the tube using the formula;
[tex] Length = \frac {2n - 1}{4} * wavelength [/tex]
Substituting into the formula, we have;
[tex] Length = \frac {2*1 - 1}{4} * 0.784 [/tex]
[tex] Length = \frac {2 - 1}{4} * 0.784 [/tex]
[tex] Length = \frac {1}{4} * 0.784 [/tex]
Length = 3.136 meters
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?
The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the chair/rocket would be ?
(80m/s/s)(70kg)=5600N
(80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
(80m/s/s)(70kg)=5600N - Fg = 5600N+(70kg)(9.8N/kg)~5600N-700N=4900N
I need the time
please explain need this ASAP
I assume you're talking about a pilot. If the ejection seat has an acceleration of 8g, then it would exert a normal force of 8g (70 kg) ≈ 5600 N.
(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)
To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the net force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)g ≈ 4900 N.
If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the net force would change to -5600 N - (70 kg)g ≈ -6300 N
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
What do we mean by a penetrative pass in the game of football?
Answer:
Penetration means forward passes can go through the opposition lines. Once these penetrative passes get through each line, it eliminates the line of players it broke through and leaves the player in possession closer to the opposition goal.
Explanation:
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
What must a scientist do in order to develop a testable hypothesis?
A. Identify a conclusion that provides the most popular explanation
for the question.
B. Determine whether experimental observations can provide
evidence to support a conclusion.
C. Copy the opinions of other scientists with similar questions.
D. Survey the preferences of other scientists who have done similar
research.
The correct answer is option B. Determine whether experimental observations can provide evidence to support a conclusion.
1. Identify the research question: Begin by clearly defining the question or problem that you want to investigate. This question should be specific and focused, allowing for clear hypotheses to be developed.
2. Conduct background research: Familiarize yourself with existing knowledge and previous research related to your question. This step will help you understand the current state of knowledge in the field and identify any gaps or areas that require further investigation.
3. Formulate a hypothesis: Based on your background research, develop a hypothesis that proposes a possible explanation or answer to your research question. A hypothesis should be a clear statement that can be tested through experimentation or observation. It should be specific, measurable, and falsifiable, meaning that it can be proven wrong if the evidence does not support it.
4. Design experiments or observations: Once you have formulated a hypothesis, consider the experimental or observational methods that can be used to test it. Determine what data you need to collect, what variables you need to manipulate or measure, and what controls should be put in place to ensure the validity of your results.
5. Predict outcomes: Based on your hypothesis, make predictions about the expected outcomes of your experiments or observations. These predictions should be derived from your hypothesis and should be specific enough to be tested.
6. Conduct the experiment or observation: Carry out the planned experiments or observations, ensuring that you collect and record data accurately. Implement proper controls and procedures to minimize any biases or confounding factors that could affect the results.
7. Analyze the data: Once you have collected the data, analyze it using appropriate statistical or analytical methods. Evaluate whether the data supports or contradicts your hypothesis and predictions.
8. Draw conclusions: Based on the analysis of your data, draw conclusions about whether the evidence supports or refutes your hypothesis. Clearly state the findings and discuss their implications in the context of the research question.
9. Communicate the results: Share your findings with the scientific community through scientific papers, presentations, or other appropriate means. Allow other scientists to review and replicate your work, fostering further discussion and advancement in the field.
Remember, the scientific process is iterative, and developing a hypothesis is just the starting point. Scientists continuously refine and revise their hypotheses based on new evidence and insights gained from their experiments and observations.
For more such questions on observations , click on:
https://brainly.com/question/16846001
#SPJ8
The angle between the reflected ray and the normal line is called the________________
Answer:
Angle of reflection
Explanation:
The angle between the reflected ray and the normal line is called the angle of reflection
Answer:
Activity 2 Directions: Solve the following problems using Charles' Law,
1. A 3.8 mL of a gas in a container has a temperature of 2.0K. What would the volume be if the temperature becomes 8.0K?
2. If at 27°C the volume of a dry gas is 3.0 ml. What is the new temperature if the volume increased to 7.5 ml?
3. The temperature of the gas is 460,0 mL at 3.5°C. Calculate the new volume at 1.5°C
