Answer:
[tex]\mu=0.185[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=17kg[/tex]
Force [tex]F=33N[/tex]
Velocity [tex]v=1.6m/s[/tex]
Distance [tex]d= 9.8m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=\triangle K.E+\triangle P.E[/tex]
Where
[tex]\triangle K.E=(F-F_f)*2[/tex]
[tex]F_f=F+\frac{\triangle K.E}{d}[/tex]
[tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]
[tex]F_f=30.8N[/tex]
Since
[tex]f = \mu*m*g[/tex]
[tex]\mu= 30.8/(m*g)[/tex]
[tex]\mu= 30.8/(17*9.81)[/tex]
[tex]\mu=0.185[/tex]
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm lần so với
trong chân không
Answer:
B.
Explanation:
sana makatulong sayo
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
To solve more questions on Work, Energy and Power, visit the link below -
https://brainly.com/question/208670
#SPJ2
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
The object has a mass of 100kg. The Tension is 200N[U]. What is the acceleration of this elevator? *
A) 8m/s/s
B) 8m/s/s[D]
C) 9.8m/s/s[D]
D) 0.5m/s/s[D]
Answer:
So the answer is B. A is wrong because negative answer = deceleration
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
Also read it;
https://brainly.com/question/11049671
#SPJ2
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
show your work please
Answer:
the horizontal component of the force is 50 N
Explanation:
Given;
force applied by the man, F = 100 N
angle of inclination of the force, θ = 60⁰
mass of the dog, m = 20 kg
The horizontal component of the force is calculated as;
[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]
Therefore, the horizontal component of the force is 50 N
Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
Its Acceleration during the upward Journey ?
If ATM is 102 kPa, what force does the atmosphere exert on the palm of your hand which has an area of 0.016 meters?
Answer:
Force = 1.632 Newton
Explanation:
Given the following data;
Pressure = 102 kPa
Area = 0.016 m²
To find what force the atmosphere exert on the palm of your hand;
Mathematically, pressure is given by the formula;
[tex] Pressure = \frac {Force}{area} [/tex]
Force = 102 * 0.016
Force = 1.632 Newton
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?
Answer:
The efficiency of Carnot's heat engine is 26.8 %.
Explanation:
Temperature of hot reservoir, TH = 100 degree C = 373 K
temperature of cold reservoir, Tc = 0 degree C = 273 K
The efficiency of Carnot's heat engine is
[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]
The efficiency of Carnot's heat engine is 26.8 %.
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
