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3 years ago

If a vehicle accelerating at 2.7 m/s2what is it's velocity at 20 meters 12.63m/s,1.03m/s,10.39m/s,6.39m/s

Physics

1 answer:

netineya [11]3 years ago

5 0

Answer:

The final velocity of the vehicle is 10.39 m/s.

Explanation:

Given;

acceleration of the vehicle, a = 2.7 m/s²

distance moved by the vehicle, d = 20 m

The final velocity of the vehicle is calculated using the following kinematic equation;

v² = u² + 2ah

v² = 0 + 2 x 2.7 x 20

v² = 108

v = √108

v = 10.39 m/s

Therefore, the final velocity of the vehicle is 10.39 m/s.

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A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bulle

vodka [1.7K]

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

$(M+ m)gh = \dfrac{1}{2}(M+m)v^2$

$gh = \dfrac{1}{2}v^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.81\times 0.31}$

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u = (M+m)v

0.026 x u = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.

5 0

3 years ago

A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb

BaLLatris [955]

By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

Initial velocity U = 0

Final velocity V = 24.3 m/s

Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/ $s^{2}$

Third Equation of motion

$V^{2} = U^{2} + 2aS$

Substitute all the necessary parameters

$24.3^{2}$ = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2 $\pi$ r

Circumference = 2 x 3.143 x 0.401

Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

4 0

2 years ago

The average depth of Indian Ocean is about 3000m.Calculate the fractional compression. ▲v/v, of water at the bottom of the ocean

Klio2033 [76]

Answer:

∴ fractional compression = 1.34 × 10⁻²

Explanation:

given,

depth of Indian ocean = 3000 m

Bulk modulus of the water = 2.2 x 10⁹ N/m²

We know,

P = P₀ + ρgh

P₀ is the atmospheric pressure

P₀ = 10⁵ N/m²

ρ is the density of the water, 1000 Kg/m³

P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²

using formula,

B = P/{-∆V/V}

B is bulk modulus and { -∆V/V} is the fractional compression

$\dfrac{-\Delta V}{V} = \dfrac{2.94 \times 10^7}{2.2 \times 10^9}$

$\dfrac{-\Delta V}{V} =1.34 \times 10^{-2}$

∴ fractional compression = 1.34 × 10⁻²

7 0

3 years ago