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Elan Coil [88]
3 years ago
14

A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight

Physics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

There is no work done.

Explanation:

<u>Given the following data;</u>

Mass = 10 kg

To find the work done?

In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.

Mathematically, work done is given by the formula;

Work done = force * distance

W = F * d

Where,

  • W is the work done
  • F represents the force acting on a body.
  • d represents the distance covered by the body.

<em>However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done. </em>

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The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

F = \frac{mv}{t}

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

F = \frac{0.07 \times 56}{0.04}

F = \frac{3.92}{0.04}

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

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3 years ago
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Sound will travel slowest through which medium?
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What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
4 years ago
PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY
m_a_m_a [10]

Hello,

<u>Solution for A:</u>

Force = 3.00N

Mass = 0.50 Kgs

Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2

<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

Mass = 0.50Kg

Time = 3.00 Seconds

Again; F = MA (Where F is force, M is mass and A is acceleration)

1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

Velocity = Acceleration * Time = 2 * 3 = 6 m/s

3 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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