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AleksandrR [38]
3 years ago
9

A bowler lifts his bowling ball a distance of 2 meters using 50 joules of energy. If, on Earth, a 1.0 kilograms mass weighs 9.8

Newton's, the mass of the bowling ball used by this bowler is about:​
Physics
2 answers:
kkurt [141]3 years ago
7 0

Answer:e

Explanation:

e

NemiM [27]3 years ago
4 0

Answer:

<em>m=2.551 kg</em>

Explanation:

Gravitational Potential Energy (GPE)

Energy is the capacity that a body has to do work due to any of its characteristics as speed, height, temperature, and many other physical magnitudes. The GPE is the one that depends on the relative height h that a body of mass m has respect to a given reference, and can be computed as

GPE=m.g.h

The bowler uses 50 Joules of energy to lift the ball by 2 meters, we can know the mass of the ball by solving the above equation for m

\displaystyle m=\frac{GPE}{g.h}

\displaystyle m=\frac{50}{(9.8).(2)}=2.551\ kg

\boxed{m=2.551\ kg}

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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