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san4es73 [151]
3 years ago
14

What is the kinetic energy of a 1200 kg roller coaster that is moving with a speed of 24.0 m/s. ?

Physics
1 answer:
Vanyuwa [196]3 years ago
3 0

Answer: 345600 J

Explanation: solution attached

KE= 1/2 mv²

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A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.
Katarina [22]
Option B would be right one
according to momentum conservation
6600*2 = 13200kgm/s
5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v
5 0
3 years ago
Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10
nydimaria [60]

Answer:

Explanation:

The formula for this is

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2} and moving things around to solve for r:

r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} } Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

8 0
3 years ago
Need an answer pls!!!<br><br><br><br><br><br> will give brainliest
solmaris [256]

Answer:

i think your answer is C

Explanation:

8 0
3 years ago
Read 2 more answers
A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate
frosja888 [35]

Answer:

N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons

Explanation:

The total charge is distributed over the two objects:

Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\

The plate and the rod must have Q_{total}/2\\. So the charge transferred from the plate to the rod is:

Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\

Number of electrons:

N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons

4 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
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