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3 years ago

Solve for the dog's displacement during segments A and B (0-20s)?

Physics

2 answers:

Katena32 [7]3 years ago

5 0

The dog walked 5 meters backwards in the first 10 seconds, then 5 meters forward from 10-20 seconds. After 20 sec he was back where he started. Displacement (20sec) = zero.

expeople1 [14]3 years ago

3 0

Explanation:

To find displacement using velocity - time graph, we need to find area under the curve during time period 0 to 20 seconds.

Displacement [ area of Triangle (b=10 units & h=1 units) - area of trapezium (sum of parallel sides=30 units & h=1 unit)

= (1/2 * 10 * 1) - (1/2 * 30 * 1)

= - 10 m

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Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers

Where will the spacecraft be when the gravitational forces acting on it are equal?

klio [65]

It would be not be able to move yet it would be in the air

6 0

3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

3 years ago

A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0

3 years ago

What are 7 examples of potential energy

yaroslaw [1]

Answer:

Hewo Otaku Kun Here! (UwU)

Explanation:

1. A rock sitting at the edge of a cliff has potential energy. If the rock falls, the potential energy will be converted to kinetic energy.

2. Tree branches high up in a tree have potential energy because they can fall to the ground.

3. A stick of dynamite has chemical potential energy that would be released when the activation energy from the fuse comes into contact with the chemicals.

4. The food we eat has chemical potential energy because as our body digests it, it provides us with energy for basic metabolism.

5. A stretched spring in a pinball machine has elastic potential energy and can move the steel ball when released.

6. When a crane swings a wrecking ball up to a certain height, it gains more potential energy and has the ability to crash through buildings.

7. A set of double "A" batteries in a remote control car possess chemical potential energy which can supply electricity to run the car.

happy to help!

from: Otaku Kun ^^

8 0

4 years ago