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Sav [38]
2 years ago
12

Please view image attached!

Physics
1 answer:
Marat540 [252]2 years ago
8 0

Answer:

you would expect a change in the graph

Explanation:

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
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SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

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What caused the slope to change around 1905 and 1906?
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Hydro then electric or solar power
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The melting point of pure water is _____.<br><br> 32°C<br><br> 100°C<br><br> 0°C<br><br> 212°C
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A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular
slega [8]

Hi there!

Angular momentum is equivalent to:

\large\boxed{L = I\omega}

L = angular momentum (kgm²/s)

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

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8 0
2 years ago
By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

m_1=43.7 kg

m_2=12.1 kg

g=9.8 m/s^2

a=\frac{1}{2}(9.8)=4.9 m/s^2

We have to find the mass of the pulley.

According to question

T_2-m_2 g=m_2 a

T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N

T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N

Moment of inertia of pulley=I=\frac{1}{2}Mr^2

(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)

Where \alpha=\frac{a}{r}

(177.87-214.13)=-\frac{1}{2}(4.9)M

-36.26=-\frac{1}{2}(4.9)M

M=\frac{36.26\times 2}{4.9}=14.8 kg

Hence, the mass of the pulley=14.8 kg

6 0
3 years ago
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