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mario62 [17]
3 years ago
12

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

48.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.50 m/s. Determine the work done (in J) by the child as the tricycle travels down the incline.
Physics
1 answer:
goblinko [34]3 years ago
5 0

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

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OlgaM077 [116]
Well if you had either the velocity or distance traveled i could tell you. But since you haven't all i can say for sure is that the water slowed the bullet down to 13m/s so lets say you knew the distance you would calculate how many meters it traveled and you would have your answer because in this situation, meters (height) =how many seconds spent going into the air. 
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3 years ago
A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min
Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

v_1=89km/h

t_1=30min

v_2=100km/h

t_2=12min

v_3=40km/h

t_3=45 min

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute=\frac{102}{60}=1.7 hour

1 hour=60 minutes

Distance=speed\times time

d_1=v_1\times t_1=80\times\frac{30}{60}=40km

d_2=100\times \frac{12}{60}=20 km

d_3=40\times \frac{45}{60}=30 km

Total distance=d_1+d_2+d_3=40+20+30=90km

Average speed=\frac{total\;speed}{total\;time}

Using the formula

Average speed=\frac{90}{1.7}=52.9Km/h

b.Total distance between the initial and final city lies along the route=90 km

4 0
3 years ago
When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

The angular velocity can be calculated as:

v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

8 0
3 years ago
A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do
ollegr [7]

Answer:

b. electric potential energy.

Explanation:

The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.

8 0
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How far would a jet going 155 m/s travel in 9 s?
docker41 [41]

Answer:

1,395 m

Explanation:

155×9

multiply m/s by 9s

5 0
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