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natulia [17]
4 years ago
14

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number.
Physics
2 answers:
Anna71 [15]4 years ago
6 0
The Gravitationa potential energy of the mass (PEG) is given by:
U=mgh
where
m is the mass
g is the gravitational acceleration
h is the heigth of the mass above the reference level (the ground)

In this problem, m=250 kg and h=0.5 m, therefore the gravitational potential energy of the mass is:
U=mgh=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J
emmainna [20.7K]4 years ago
3 0

Answer:

i can confirm it is 1,225

Explanation:

based on the website. I just did this question.

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A ferris wheel car with a mass of 350 kg, travels in a
SpyIntel [72]

Answer:Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = ac. Thus, the magnitude of centripetal force Fc is Fc = mac.

By using the expressions for centripetal acceleration ac from  

a

c

=

v

2

r

;

a

c

=

r

ω

2

, we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:  

F

c

=

m

v

2

r

;

F

c

=

m

r

ω

2

.

You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r, you get  

r

=

m

v

2

F

c

.

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.

Figure 1. The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the Fc, the smaller the radius of curvature r and the sharper the curve. The second curve has the same v, but a larger Fc produces a smaller r′.

Explanation:

6 0
3 years ago
_____ is the densest planet. Mercury Venus Earth Mars
Orlov [11]
The answer is Venus, with the density of 5.204 grams per cubic centimeter
5 0
3 years ago
5.
Kamila [148]

Answer:  (x - 4)² + (y - 7)² = 9

<u>Explanation:</u>

The equation of a circle is: (x - h)² + (y - k)² = r²    where

  • (h, k) is the center
  • r is the radius

Given: (h, k) = (4, 7)

Find the intersection of the given equation and the perpendicular passing through (4, 7).

3x - 4y = -1

     -4y = -3x - 1

        y=\dfrac{3}{4}x-1

              m=\dfrac{3}{4}      -->      m_{\perp}=-\dfrac{4}{3}

y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}

Use substitution to find the point of intersection:

x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6

Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius

r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3

Input h = 4, k = 7, and r = 3 into the circle equation:

(x - 4)² + (y - 7)² = 3²

(x - 4)² + (y - 7)² = 9

4 0
3 years ago
When sediment created by weathering and moved by erosion settles in a different location due to gravity
Stolb23 [73]

Answer:

Deposition

Explanation:

3 0
3 years ago
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Oduvanchick [21]
I really don’t know I just need some points to help me though this test I’m very sorry
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