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

#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 x away from the wall at time t according to

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

so that after t = 8 s, it will be

x = 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