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Llana [10]
3 years ago
15

A biker starts moving from rest with a constant acceleration of 2.0 m/s^2. What is his velocity after 5.0 s?

Physics
1 answer:
Ber [7]3 years ago
5 0

a = v-u/t

2 = v - 0 / 5

2 = v/5

v = 10 m /s

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P
worty [1.4K]

Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure P_{in}= 1.90\times10^{5}\ Pa

Area A=3.90\times10^{-4}\ m^2

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

F_{net}=F_{in}-F_{out}

F_{net}=(P_{in}-P_{out})A

Where, P_{in} =inside Pressure

P_{out} =outside Pressure

A = area

Put the value into the formula

F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}

F_{net}=34.59\ N

Hence, The magnitude of the force is 34.59 N.

6 0
3 years ago
A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera
lana66690 [7]

Answer:

420000N

Explanation:

Given parameters:

Mass of the train  = 5.6 x 10⁵kg

Acceleration  = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

   Force  = mass x acceleration

Resultant force that acts on the train is given below;

 Force  = 5.6 x 10⁵kg  x 0.75m/s²  = 420000N

7 0
3 years ago
HELP<br><br> These images represent waves. <br><br> Which wave has the highest frequency?
Ierofanga [76]

Wave C

because it is the closest or each wave is closer than the others

7 0
3 years ago
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nignag [31]

Answer:

B

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7 0
2 years ago
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
3 years ago
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