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Mademuasel [1]
2 years ago
5

If you lift weights faster than normal you have increased what?

Physics
1 answer:
egoroff_w [7]2 years ago
8 0
I belive it would power I’m not 100% sure. If it’s not power then force
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The movement of water is able to transport minerals and nutrients. Which statement best explains why water is able to do this?
Marrrta [24]

no it can't do this why because I think that it is water and it can not go any where.

3 0
3 years ago
Read 2 more answers
1. Any push or pull is called
timama [110]

Answer:

b  ok

Explanation:

7 0
3 years ago
A CAR ACCElerates FROM REST To
Ilia_Sergeevich [38]

Answer:

<u>666.6 kW</u>

Explanation:

<u>Power Formula</u>

  • Power = Force × Velocity

<u>Calculating Force</u>

  • F = ma
  • F = m(v - u / t) [From the equation v = u + at]
  • F = 2,000 (50/7.5)
  • F = 2,000 (20/3)
  • F = 40,000/3
  • F = 13333.3 N

<u>Solving for Power</u>

  • P = 13333.3 × 50
  • P = 666666.6 W
  • P = <u>666.6 kW</u>
5 0
2 years ago
A 20 kg bike accelerates at 10 m/s2. What was the force?
agasfer [191]

Answer:

200 N = 200 Newtons

Explanation:

Just use the formula F = m*a

F = Force in Newtons

m = mass and is 20 kg

a = acceleration and is 10 m/s^2

F = 20 * 10

F = 200 Newtons.

7 0
3 years ago
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
3 years ago
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