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Fynjy0 [20]
3 years ago
11

Farah claps his 600times in 60seconds. what is the friquency and periodb​

Physics
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

\huge\boxed{\sf Frequency = 10\ Hz}

\huge\boxed{\sf Time\ Period = 0.1\ secs}

Explanation:

<u>Given:</u>

No. of times = 600 times

Time = 60 seconds

<u>Required:</u>

Frequency = f = ?

Time Period = T = ?

<u>Solution:</u>

<h2>Frequency = No. of Times / Time</h2>

Frequency = 600 / 60

<u>Frequency = 10 Hz</u>

<h2>Time Period = 1 / Frequency</h2>

Time Period = 1 / 10

<u>Time Period = 0.1 secs</u>

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
★CHECK MY ANSWER PLEASE★
Sergeu [11.5K]
You are correct!
Happy to assist you!
8 0
3 years ago
Read 2 more answers
Which factors affect friction between two solid surfaces?
a_sh-v [17]
1) newton of the surfaces in contact (the smoothness of the surfaces)

2) how hard the surfaces press together
6 0
3 years ago
Read 2 more answers
A 5kg ball is on top of the school building at a height of 40m above the ground.
mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0
2 years ago
A coil with 55 loops has its flux change from 0 Wb to 0.0266 Wb in a certain amount of time, generating 5.22 V of EMF. How much
Eduardwww [97]

Answer:

0.280 s

Explanation:

I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.

4 0
3 years ago
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