Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2

A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t

zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0

2 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

2 years ago

A car is moving with a speed of 22 m/s. The driver then brakes, and the car comes to a halt after 6.5 s. What is the distance co

Shalnov [3]

The answer would be 72 miles. Hope this helped.

3 0

3 years ago

A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i

inna [77]

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

7 0

2 years ago

For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit

kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2; P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ = cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

8 0

2 years ago