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3 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s2. how long will it take for that object to travel 3400 m?

Physics

1 answer:

Montano1993 [528]3 years ago

4 0

15 seconds i may be wrong though

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Which of the three recording stations was closest to the epicenter of the earthquake, based on the seismograms shown below?

Nastasia [14]

Answer:C

Explanation:took quiz

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3 years ago

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a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after

STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

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3 years ago

What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why

Veseljchak [2.6K]

Answer:

it creates a gas called carbon dioxide. The gas begins to expand in the bottle and starts to inflate the balloon

Explanation:

Why does this happen? well, The faster-moving particles inside the bottle start to move faster and faster and soon they expand to fill the balloon.

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2 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

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3 years ago

Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?

Naya [18.7K]

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

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3 years ago