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wariber [46]
3 years ago
12

A 3.0-kilogram toy dump truck moving with a speed of 4.0 m/s starts up a ramp. How high does the truck roll before it stops?

Physics
1 answer:
baherus [9]3 years ago
7 0

Before it reaches the ramp, the toy's kinetic energy is

                                        (1/2) (mass) (speed²)

                                   =   (1/2) (3 kg) (16 m²/s²)

                                   =          24 Joules .

On the way up the ramp, its kinetic energy will change to
gravitational potential energy, and it'll stop when it's height
gives it  24 joules of potential energy.

         Potential energy  =  (mass) (gravity) (height)

                    24  joules  =  (3 kg) (9.8 m/s²) (height)

Divide each side by  29.4 kg-m/s²  (that's the same as 29.4 newtons):

                    Height  =  24 joules / 29.4 newtons

                                =      0.816 meter higher than the bottom of the ramp.
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At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan
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Answer:

2.67kg

Explanation:

The maximum velocity, v _ {max} of a body experiencing simple harmonic motion is given by equation (1);

v_{max}=\omega A............(1)

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\omega=\sqrt{\frac{k}{m}}.................(2)

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We can make \omega the subject of formula in equation (1) as follows;

\omega=\frac{v_{max}}{A}.................(3)

We then combine equations (2) and (3) as follows;

\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)

According to the problem, the following  are given;

v_ {max }=1.92m/s\\A=0.21m\\k=223N/m

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}

9.143=\sqrt{\frac{223}{m}}

Squaring both sides, we obtain the following;

9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg

8 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
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part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
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