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3 years ago

What is the quantity that has the units of m/s^2?

Physics

1 answer:

Luden [163]3 years ago

8 0

Answer:

What is the quantity that has the units of m/s²?

Velocity
Acceleration✓
Force
Momentum

Acceleration is the quantity that has the units of m/s².

Explanation:

Acceleration is the rate of change of velocity.

Acceleration=(Final velocity-Initial velocity)/Time.

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An airplane increases its velocity from 60m/s to 80m/s in a time of 10s.

Elenna [48]

$\huge\fbox{Answer ☘}$

$acceleration \: = \frac{v - u}{t} \\ = > \frac{80 - 60}{10} \\ \\ = > \frac{20}{10} \\ \\ = > 2m \: s {}^{ - 2}$

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2 years ago

What current is produced with a voltage of 6.0 V and a resistance of 445 ohms?17.4 mA

san4es73 [151]

Answer:

i Believe its 17.4 mA

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2 years ago

A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

= 1000 (0 - 20)

= 1000 (-20)

= -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

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3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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3 years ago

If we increase the force applied to an object and all other factors remain the same that amount of work will

worty [1.4K]

Hello there.

If we increase the force applied to an object and all other factors remain the same that amount of work will

C. Increase

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3 years ago

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