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abruzzese [7]
3 years ago
13

A 120 -kg crate accelerates toward the positive x-direction . If the magnitude of the force due to friction is 74.4 N , what for

ce must pull on the crate to give it a 2.5 m/s ^ s acceleration ?
Physics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

The applied force to accelerate the crate is 374.4 N

Explanation:

Given;

mass of the crate, m = 120 kg

magnitude of force due to friction, Fk = 74.4N

acceleration of the crate, a = 2.5 m/s²

The net horizontal force on the crate is calculated as;

∑Fx = F - Fk

ma = F - Fk

F = ma + Fk

where;

F is the applied force to accelerate the crate by 2.5 m/s²

F = (120 x 2.5) + (74.4 N)

F = 300 N + 74.4 N

F = 374.4 N

Therefore, the applied force to accelerate the crate is 374.4 N

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It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of f
Anestetic [448]

Answer:

A = 8.34 x 10^(5) m²

Explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

P_rad = 2I/c

Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²

7 0
3 years ago
A uniform electric field of magnitude 375 n/c pointing in the positive x - direction acts on an electron, which is initially at
Finger [1]
(a) The force exerted by the electric field on the electron is given by the product between the electron charge q and the intensity of the electric field E:
F=qE=(1.6 \cdot 10^{-19}C)(375 N/C)=6\cdot 10^{-17}N
Under the action of this force, the electron moves by:
\Delta x = 3.20 cm=0.032 m
And the work done by the electric field on the electron is equal to the product between the magnitude of the force and the displacement of the electron. The sign has to be taken as positive, because the direction of the force is the same as the displacement of the electron, so:
W=F \Delta x= (6\cdot 10^{-17}N)(0.032 m)=1.9 \cdot 10^{-18}J

(b) The electron is initially at rest and it starts to move under the action of the electric field. This means that as it moves, it acquires kinetic energy and it loses potential energy. The change in potential energy is the opposite of the work done by the electric field:
\Delta U = U_f - U_i = -1.9 \cdot 10^{-18} J
Where Uf and Ui are the final and initial potential energy of the electron.

(c) For the conservation of energy, the sum of the kinetic energy and potential energy of the electron at the beginning of the motion and at the end must be equal:
U_i + K_i = U_f + K_f (1)
where Ki and Kf are the initial and final kinetic energies.
The electron is initially at rest, so Ki =0, and we can rewrite (1) as 
U_i - U_f = - \Delta U = K_f = \frac{1}{2}m_e v_f^2
and by using the mass of the electron me, we can find the value of the final velocity of the electron:
v_f= \sqrt{ -\frac{2 \Delta U}{m_e} }= \sqrt{- \frac{2(-1.9 \cdot 10^{-18} J)}{9.1 \cdot 10^{-31} kg} } =2.04 \cdot 10^6 m/s



7 0
3 years ago
Two of the angles in a triangle are complementary. The third angle is twice the measure of one of the complementary angles. What
Marrrta [24]

Explanation:

Let the 2 be X and Y.

Z is 3rd angle.

Z=2X

w.k.t Sum of angles in a triangle is 180°

Z+X+Y=180°

2X+90°=180°

2X=90°

=>X=45°

=>Y=90°-45°=45°

=>Z=2X=2(45°)=90°

Hope this helps you.

6 0
3 years ago
The football player running towards the goal line has
blagie [28]

the football player has speed

8 0
3 years ago
Read 2 more answers
Need help with these two questions Please?
Ksivusya [100]
Im too young for physics, sorry!
5 0
3 years ago
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