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3 years ago

A bowling ball travels at 2.0 m/s. It has 16 J of kinetic energy. what Is the mass of the bowling ball in kilograms ?

Physics

1 answer:

zaharov [31]3 years ago

5 0

Kinetic energy = 1/2 × m × v^2
16 = 1/2 × m × 2^2
16 = 1/2 × m × 4
16 = 2 × m
16/2 = m
8 = m

so the mass is 8 kg

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

3 years ago

A person starts at a position of 16 meters and finishes at a position of

DochEvi [55]

Answer:

38 is a good girl and a great place to work for u and I miss you y and I miss you much love your love and love to

7 0

2 years ago

The heat required to melt a piece of copper is (82 x 10 ^5 J). The heat of fusion of copper is (2.05×10 ^5 J/ kg). What is the m

Musya8 [376]

Answer:

Explanation:

Givens

Heat of Fusion = 2.05 * 10^5 J / kg watch the units.

Heat to actually melt the copper = 82 10^5 J

Formula

Mass of copper = Heat / Heat of Fusion

Solution

Mass of copper = 82*10^5 J / (2.05 * 10^5 J / kg)

Mass of copper = 40 kg

Notice that the kg is in the denominator of the second fraction. The rules of fractions would tell you the 1/1 / / 1 /kg . You take the right fraction and turn it upside down and multiply. 1 / 1 * kg/1 = 1* kg / 1*1 which is just kg.

Answer 40 kg of copper

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2 years ago

How do I find the velocity of a moving object?

Aliun [14]

If you know the distance and the time I travelled that distance.

You just have to divide the time from the distance to get velocity

V =d
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t

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3 years ago

Fifty (50) grams of vinegar and (2) grams of baking soda are mixed together. A bubbling reaction takes place. The mixture is wei

Ludmilka [50]

Answer:

the gas that escaped from the mixture contained the missing weight

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2 years ago