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swat32
3 years ago
5

A car accelerates uniformly from rest at 2.2 m / s^2 for 3.0 s.Calculate the speed of the car at time t = 3.0 s

Physics
1 answer:
Alexxx [7]3 years ago
4 0

Answer:

6.6m/s

Explanation:

we know that

v = u + at

= 0+ 2.2*3

=6.6m/s

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and c
SVEN [57.7K]

Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s

Given:

v₀ = 85.1 ft/s

v = 0 ft/s

t = 0.24 s

Find: a

v = at + v₀

a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

a = -354 ft/s²

Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

7 0
3 years ago
When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl
zimovet [89]
Static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
           Cκ = 0.0822 
3 0
3 years ago
Does the earths magnetic field change with time?
ZanzabumX [31]

Answer:

As per the fossil fuel records, magnetic field reversal does not impact living beings. It will take almost a century for the poles to complete the shift. Meanwhile, the earth is left with almost zero magnetic field.

6 0
2 years ago
A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0
2 years ago
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