How much power will it take to move a 50Kg box 10m across a floor that has -50N of Frictional forces in 3 seconds?

Physics

1 answer:

likoan [24]3 years ago

8 0

Answer: 83.3 W

Explanation: I think, I’m not sure. If I’m wrong correct me ;)

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R'=2 Ohm R"=1,5 Ohm R"'=2 Ohm

Andrew [12]

Answer:

2.5 ohm

Explanation:

R' and R''' are parallel

So,

1/R1= 1/R' + 1/R'''

1/R1 = 1/2 + 1/2

1/R1 = 1

so,

R1= 1 ohm

Now R1 and R'' are in series

so,

R= R1 + R''

R= 1 + 1.5

R= 2.5 ohm

5 0

3 years ago

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

4 years ago

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of

Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

$h_i$ = Initial height

$h_f$ = Final height

From conservation of Energy

$\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}$