PHYSICS COLLEGE

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer 1
Answer:

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

F=(mv^2)/(r)

v=\sqrt{(Fr)/(m)}

v=\sqrt{(0.025* 0.29)/(0.01)}

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

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The digital exchange of structured data is called ?

Answers

Answer:

Electronic data interchange

A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\nor\nP=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_(1)*v_(1))+F*t=(m_(1)*v_(2))

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

(10*1)+F*5=10*4\n10+5*F=40\n5*F=40-10\n5*F=30\nF=6[N]

Nitrogen makes up about what percent of a human's body weight?

Answers

Answer:

the answer is 3.3 %

Explanation:

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?


A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

          \text { Work done }=\text { Force } * \text { displacement }

          \text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}

It is known that 1 N=1 \mathrm{kg} \mathrm{ms}^(-2)

So, the work done can be expressed in k g m s^(-2) as,

         \text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)

It is known that 1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2), so the conversion of units from Nm to erg will be done as follows:

\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)

Thus, work done in ergs is 3200 ergs.

What is a light year​

Answers

Answer:

A light-year is the distance light travels in one year.

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).
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