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OleMash [197]
2 years ago
10

The roller coaster from problem #1 then tops a second hill at 15.0 m/s, how high is the second hill?

Physics
1 answer:
navik [9.2K]2 years ago
4 0

Answer:

68.5

Explanation:

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#2
sladkih [1.3K]

Answer:

mass and distance

Explanation:

force is mass while motion can also be regard as distance or movement

3 0
3 years ago
A mass of 100 kg is pulled by a 392 N force in the +X direction along a rough surface (uk=0.4) with uniform velocity v=20 m/s. W
alexdok [17]

Answer:

The total work done will be zero.

Explanation:

Given that,

Mass = 100 kg

Force = 392 N

Velocity = 20 m/s

Distance s= 10 m

We need to calculate the work done

Using balance equation

The net force will be

F'=F-\mu mg

F'=392-0.4\times100\times9.8

F'=0

The net force is zero.

Hence, The total work done will be zero by all forces on the object.

5 0
2 years ago
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant condit
Darya [45]
Answer hopefully = C. self-efficacy
4 0
2 years ago
Read 2 more answers
A bicycle is 10 m from a wall traveling at a rate of 4 m/s. How far from the wall will it be if it accelerates at a rate of 12 (
kompoz [17]

The bicycle is a distance <em>x</em> away from the wall at time <em>t</em> according to

<em>x</em> = 10 m + (4 m/s) <em>t</em> + 1/2 (12 m/s²) <em>t</em> ²

so that after <em>t</em> = 8 s, it will be

<em>x</em> = 10 m + (4 m/s) (8 s) + 1/2 (12 m/s²) (8 s)² = 426 m

away from the wall.

5 0
3 years ago
A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the aver
nevsk [136]

Answer:

When he weight of the car is 8.55 x 10^{3} N then power = 314.012 KW

When he weight of the car is 1.10 x 10^{4}  N then power =  43.76 KW

Explanation:

Given that

Initial velocity V_{1} = 0

Final velocity V_{2} = 24.8 \frac{m}{s}

Time = 7.88 sec

We know that power required to accelerate the car is given by

P = \frac{change \ in \ kinetic \ energy}{time}

Change in kinetic energy Δ K.E = \frac{1}{2} m (V_{2}^{2} - V_{1}^{2}   )

Since Initial velocity V_{1} = 0

⇒ Δ K.E = \frac{1}{2} m V_{2}  ^{2}

⇒ Power P = \frac{1}{2} \frac{m}{t}  V_{2} ^{2}

⇒ Power P = \frac{1}{2} \frac{W}{g\ t}  V_{2} ^{2}   -------- (1)

(a). The weight of the car is 8.55 x 10^{3} N = 8550 N

Put all the values in above formula

So power P = \frac{1}{2} \frac{8550}{(9.81)\ (7.88)} (24.8) ^{2}

P = 314.012 KW

(b). The weight of the car is 1.10 x 10^{4} N = 11000 N

Put all the values in equation (1) we get

P = \frac{1}{2} \frac{11000}{(9.81)\ (7.88)} (24.8) ^{2}

P = 43.76 KW

5 0
3 years ago
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