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3 years ago

An object, accelerating from rest at a constant rate, travels over 28 m in 11 s. What is its final velocity?

Physics

1 answer:

Paul [167]3 years ago

4 0

Answer:

answer is 3

Explanation:

by using s= [(v+ u)/2] x t

28= (v+ 0)/2 x 11

v= 5.09 ms^-1

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A charge of 3 nC is uniformly distributed around a ring of radius 10 cm that has its center at the origin and its axis along the

Step2247 [10]

Answer:

$W=2.592*10^{-7} J$

Explanation:

GIVEN DATA:

Charge 3 nC

Radius of ring is 10 cm

A point charge 1 nC

Distance of point charge is 2.5 m

We know that voltage is calculated as

$V(0) =\frac{KQ}{r}$

$=\frac{(9*10^{-9})(3*10^{-9})}{0.1}$

V = 270 V

At x = 2.5m

$r = \sqrt{(0.1^2+2.5^2)}$

r = 1.581 m

$V(2) = \frac{(9*10^9)(3*10^{-9})}{1.581}$

V(2) = 10.801 V

Work done is calculated as

W= q(dV)

$W = (1*10^{-9})(270 - 10.80)$

$W=2.592*10^{-7} J$

8 0

4 years ago

А)

loris [4]

Image of the plate is missing, so i have attached it.

Answer:

A) t = 7.854 mm

B) σ = 26.67 MPa

Explanation:

A) From shearing of rivet, formula for pressure is;

P = τ•A_rivets

Where;

τ is allowable stress

A_rivets is Area of rivet

We are given:

τ = 60 MPa

Diameter; d = 20 mm

A_rivets = πd²/4 = π × 20²/4 = 100π

Thus;

P = 60 × 100π

P = 6000π N

From bearing of plate material, we can calculate pressure as;

P = σ_b•A_b

We are given;

σ_b = 120 MPa

A_b is area of plate = 20t

Where t is the thickness

Thus;

6000π = 120 × 20t

t = 6000π/(120 × 20)

t = 7.854 mm

B) Largest average tensile stress is given by the formula;

σ = P/A

Where A = 110t - 20t

A = 90t

A = 90 × 7.854

Thus;

σ = 6000π/(90 × 7.854)

σ = 26.67 MPa

8 0

3 years ago

When you blink your eye, the upper lid goes from rest with your eye open to completely covering your eye in a time of 0.024 s.

dusya [7]

a) Distance moved by the top lid during a blink: 1 cm (estimate)

b) The acceleration is $34.7 m/s^2$

c) The final speed is 0.83 m/s

Explanation:

For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.

Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the eyelid, we have:

u = 0 (it starts from rest)

s = 1 cm = 0.01 m is the distance covered

t = 0.024 s is the time taken

Solving for a, we find the acceleration:

$a=\frac{2s}{t^2}=\frac{2(0.01)}{0.024^2}=34.7 m/s^2$

The final speed of the upper eyelid can be found by using another suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

For the eyelid in this problem, we have

u = 0

$a=34.7 m/s^2$

t = 0.024 s

Substituting, we find the final speed:

$v=0+(34.7)(0.024)=0.83 m/s$

Learn more about accelerated motion:

brainly.com/question/9527152

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#LearnwithBrainly

4 0

3 years ago

I know the acceleration due to gravity (ie 9.8 m/s2) will have a negative sign when falling down, a positive one when going up.

rewona [7]

Answer:

Yes

Explanation:

Accerelation is measured by change in velocity. So naturally, if an object is slowing down, its velocity is decreasing so acceleration is negative. If it is speeding up velocity is increasing so positive acceleration.

(Velocity final - Velocity initial)/t

Note that this does not apply only to gravity, but to all linear accelerations

6 0

2 years ago

How does the ringing sound of a telephone travels from the phone to your ears?

Sedaia [141]

The "ringing sound" are actually the soundwaves travelling from the phone to your ears.

6 0

3 years ago