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3 years ago

Is it possible to accelerate while moving constant speed

Physics

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

Explanation:

An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.

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Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry

USPshnik [31]

Answer is given below

Explanation:

This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle
So, most of the weight of full loaded wheelbarrow's load goes on that's wheel and due to friction force between wheel and surface it can easy to move
He uses force to rotate the wheel, much more than the force applied to the rim of the wheel on the axis of rotation or torque

3 0

3 years ago

If a car used 260.000 W of power to complete a race in 15 s. how much work did the car do?

marta [7]

Work done by the car is 3900 J

Explanation:

Power and work are related by the equation, Power = Work Done/Time
Power is the rate at which work is done.
Here, the car uses power of 260 W and time taken is 15 s.

Work Done = Power × Time

= 260 × 15 = 3900 J

3 0

3 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

3 0

3 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />

8 0

3 years ago

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago