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Galina-37 [17]
3 years ago
12

A student walks 3

Physics
1 answer:
pochemuha3 years ago
8 0

Answer: The speed of the student is 5.45km/hr.

Explanation: The important concept here is to convert 33 minutes to hours by diving by 60 since one hour is equal to 60 minutes to get 0.55hr

Then velocity is equal to distance over time.

v=d/t

v= 3km/0.55hr

v=5.45km/hr

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What is the kinetic energy of a 5-kg object moving at 7 m/s?
Sunny_sXe [5.5K]
To answer this question, we would need the formula for the kinetic energy which is Kinetic Energy = ½ x m x v^2

Where the following means: m is the mass of the object and v is the velocity

So At 7 m/s: Kinetic Energy = ½ x 5 x 7^2 = 122.5 J is the answer
6 0
3 years ago
Calculate the average speed of a car stuck in<br> traffic that drives 15 km in 3 hours?
dusya [7]

Answer:

5km/h

Explanation:

Speed= distance/time

Speed=15/3

speed=5Km/h

7 0
2 years ago
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?
Ksenya-84 [330]

Given:

I = 30dB

P = 66 × 10^{-9} Pa

Solution:

Formula used:

I = 20\log_{10}(\frac{P}{P_{o}})           (1)

where,

I = intensity of sound

P = absolute pressure

P_{o} = reference pressure

Using Eqn (1), we get:

30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}

P_{o} = \frac{66\times 10^{-9}}{10^{1.5}}

P_{o} = 2.08 × 10^{-9} Pa

4 0
3 years ago
A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa
Shalnov [3]

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, \theta=28^{\circ}

We need to find the angular speed of the sphere at the bottom of the hill if it starts  from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

I is the moment of inertia of the sphere, I=\dfrac{2}{5}mr^2

Also, v=r\omega

h is the height of the ramp, h=l\ sin\theta

mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2

On solving the above equation we get :

\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}

\omega=\sqrt{\dfrac{10\times 9.8\times 6\ sin(28)}{7(0.2)^2}}

\omega=31.39\ rad/s

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.

6 0
3 years ago
When a potential difference of 20 V is applied to a given wire, it conducts 0.25 A of current.?
Anon25 [30]
Well if the wire has an insane length or is made out of a high resistance material. But, you're probably looking for a resistor : R = U / I ; R = 20V / 0.25A
4 0
3 years ago
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