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3 years ago

Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)

Physics

1 answer:

olganol [36]3 years ago

8 0

Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)

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A car travels 50 kilometers in 30 minutes. Which about the motion of the car are true?

irga5000 [103]

Answer:

1500

Explanation:

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1 year ago

1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?

Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

Mass m = 1.75 kg

Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

$K.E=\frac{1}{2}mv^2$

where

m is the mass (kg)

v is the velocity (m/s)

K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2$

$K.E=1458\times 1.75$

$K.E=2551.5$ J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

7 0

2 years ago

A diver is is pushed upwards by a diving board. She weighs 739 N and accelerates from rest to a speed of 4.60 m/s while moving 0

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Answer:

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2 years ago

A venturi is constructed of a 10.0 cm pipe with a 2.0 cm diameter throat. Water pressure in the pipe is twice atmospheric pressu

postnew [5]

Answer:

$P_t=5066250.0696\ Pa=50\ atm$

Explanation:

Given:

diameter of pipe, $d=0.1\ m$

diameter of throat, $d_t=0.02\ m$

velocity of flow, $v=0.4\ m.s^{-1}$

Pressure in the pipe is twice the atmospheric pressure:

$P=2\times 101325=202650\ Pa$

Now the force of flow of water:

$F=P\times A$

now we find cross sectional area of the pipe:

$A=\frac{\pi.d^2}{4}$

$A=\pi \times\frac{0.1}{4}$

$A=0.007854\ m^2$

Therefore,

$F=202650\times 0.007854$

$F=1591.6094\ N$

Now the area at throat:

$A_t=\frac{\pi.d_t^2}{4}$

$A_t=\frac{\pi\times 0.02^2}{4}$

$A_t=0.000314\ m^2$

Therefore pressure at throat:

$P_t=\frac{F}{A_t}$

$P_t=\frac{1591.6094}{0.000314}$

$P_t=5066250.0696\ Pa=50\ atm$

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2 years ago

How can free runners extend the time of impact when they land on the ground after jumping long distance

SVETLANKA909090 [29]

Answer:

By leaping.

Explanation:

When free runners run, they also leap. They do this so they can get more distance and air time. Leaping will allow them to get more distance per jump.

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2 years ago