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2 years ago

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A roller coaster is released from the top of a track that is 57 m high. Find the roller coaster speed when it reaches ground lev

el.

1 answer:

Art [367]2 years ago

7 0

Hi there!

Assuming the track is frictionless:

$E_i = E_f\\\\PE = KE\\\\mgh = \frac{1}{2}mv^2$

Cancel out the masses and rearrange to solve for velocity:

$gh = \frac{1}{2}v^{2}\\\\v = \sqrt{2gh}\\\\$

Plug in the given height and let g = 9.8 m/s²:

$v = \sqrt{2(9.8)(57)} = \boxed{33.42 m/s}$

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A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 32.67 N B. 2.04 kg C. 1.96 kg D. 196 N

Brilliant_brown [7]

M=20 kg
g on earth=9.8m sec^-1
F=m*a
F=20*9.8
F=196N

7 0

3 years ago

Read 2 more answers

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

<em>radius of the curve, r = 90 m</em>
<em>angle of inclination, θ = 10.8⁰</em>
<em>speed of the car, v = 75 km/h = 20.83 m/s</em>
<em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

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8 0

3 years ago

In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.

KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m$

So, height of the jump is 3.97305 m.

4 0

3 years ago

A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If

Gala2k [10]

Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

Let g = 9.81 m/s2

$\Delta E_p = mgH - mgh = mg(H - h) = 0.32*9.81*(2 - 1.2) = 2.51 J$

Since 1.8J of 2.51 J is due to work by air resistance, the rest of the energy (2.51 - 1.8 = 0.71 J) is would go to heating in the ground and ball when it bounces.

5 0

3 years ago