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3 years ago

What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,

1 answer:

7 0

A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2

.:. The acceleration is 28.27 m/s^2

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Which lenses are convex? Check all that apply. 1 2 3 4 5 6

Goryan [66]

Answer:

1, 2, 3, and 6 are the answers.

Explanation:

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5 0

3 years ago

Read 2 more answers

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

7 0

3 years ago

A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h

BartSMP [9]

Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia

4 0

3 years ago

Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe

Vsevolod [243]

The period of the pendulum is the reciprocal of the frequency:
$T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s$

The period of the pendulum is given by
$T=2 \pi \frac{L}{g}$
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can calculate the length of the pendulum L:
$L=g \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2) \frac{(6.25 s)^2}{(2 \pi)^2}=9.71 m$

7 0

3 years ago

You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of

Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

$X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}$