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3 years ago

How much force is needed to lift a 25-kg mass at a constant verlocity?

Physics

2 answers:

Troyanec [42]3 years ago

5 0

-- In order to achieve constant verlocity, the net force on the mass must be zero. So if there ARE any forces acting on it, they must be balanced.

-- There is already a force on the mass that can't be eliminated . . . the force of gravity.

-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = 245N in the downward direction.

-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of 245N, all in the upward direction.

-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.

ehidna [41]3 years ago

4 0

Answer:

f = 245 N

Explanation:

see the solution

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If a force of 19 N is applied to a crate to push it 23 m, how much work is done on the crate by the force?

swat32

Answer:

437 Joules

Explanation:

Use the formula for work directly

(work) = (force) x (displacement)

to get

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3 years ago

A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?

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Answer:

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Explanation:

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3 years ago

A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelera

user100 [1]

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

Vf₁ = 25 m/s

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

s₂ = 90 m

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

5 0

3 years ago

A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a desk. What is the initial vertica

PSYCHO15rus [73]

Answer:

The Initial vertical velocity is zero

Explanation:

Given that,

A book is pushed with an initial horizontal velocity, Vx = 5 m/s

According to projectile motion, the horizontal component of velocity i.e., in X-axis remains the same though out its path.

The vertical velocity varies i.e., the velocity towards the Y- axis.

The velocity of the falling book in its trajectory is the resultant velocity. Due to the action of gravity, it keeps on increasing with time.

The resultant velocity is given by relation

$V_{R} = \sqrt{{V_{x}^{2}} +[ {gt}]^{2} }$

where

gt = Vy - vertical velocity

g - acceleration due to gravity

t - time

Since, the initial horizontal velocity at that instant is equal to the resultant velocity.

Vx = 5 m/s

$V_{R}$ = 5 m/s

So, on substituting in the above equation

Vy = 0 m/s

Therefore, the initial vertical velocity is equal to zero

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