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2 years ago

A 1.0 kg object is attached to a 0.50 m string. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s.

Physics

1 answer:

7nadin3 [17]2 years ago

4 0

Given that,

Mass of the object, m = 1 kg

It moves in a circle of radius 0.5 m with a speed of 5 m/s

To find,

The direction of the acceleration.

Solution,

Whenever an object moves in a circular path, the only force that acts on its is centripetal force which is given by the formula as follows :

$F=\dfrac{mv^2}{r}$

The centripetal acceleration acts in the direction of force. It acts along the radius of the circular path. In this figure, the direction of the acceleration is shown at point d.

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A ball has a mass of 2 kg and is thrown with a force of 8 Newtons for .35 seconds. What is the ball's change in

Anna [14]

Answer:

1.4s

Explanation:

Given parameters:

Mass of ball = 2kg

Force = 8N

Time = 0.35s

Unknown:

Change in velocity = ?

Solution:

To solve this problem, we use the expression obtained from Newton's second law of motion which is shown below:

Ft = m(v - u)

So;

Ft = m Δv

F is the force

t is the time

m is the mass

Δv is the change in velocity

8 x 0.35 = 2 x Δv

Δv = 1.4s

6 0

2 years ago

It is easier to open the lid of a can using a spoon why?

oksian1 [2.3K]

Answer:

It depends on how you use the spoon...

if you are keeping one end of the spoon and pressing other end, the force you provide is supported by the force due to gravity... Hence it is easy to open this way :)

F + G ... Where F is the force you provide and G is the force due to gravity.

8 0

3 years ago

Read 2 more answers

How does the position of an object relate to the energy stored in an object?

Kamila [148]

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

4 0

2 years ago

Help please!

JulsSmile [24]

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

7 0

2 years ago

A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.

Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\$

assume the bullet goes to right side and the gravitational acceleration =10 $ms^{-2}$

so now the weight of the rifle= $\frac{25}{10}$

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}$

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0

2 years ago