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3 years ago

A student lifts 100 N of books 1 m to his shoulder. He walks forward

Physics

1 answer:

Delvig [45]3 years ago

7 0

Hope this will help u.....:)

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A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio

Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

v² = u² + 2 a h

0 = 64² - 2 x 32 x h

64 h = 64²

h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0

3 years ago

In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor

VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

$I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596$ kg-m2

$I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2$

by perpendicular axis theorem

$I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2$

4 0

3 years ago

A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright

kumpel [21]

Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³

$\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}$

λ = 596 nm.

8 0

3 years ago

Brainliest if correct Question 1 of 10

Art [367]

Answer:

Explanation:

hope this helps you

4 0

2 years ago

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago