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Schach [20]
3 years ago
12

Please help don't for get to show work

Physics
1 answer:
emmasim [6.3K]3 years ago
3 0
-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

The direction is the angle whose tangent is (32/45) south of east.

tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.
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Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0
3 years ago
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
3 years ago
In the following description, QUOTE the portion that addresses weather in the area.
olga nikolaevna [1]

Answer:

Portland normally had rainy, but mild winters. Yesterday it rained 0.75 inches in Portland.

Explanation:

This talks about the weather

4 0
3 years ago
Can somebody help me with this quiz please
mezya [45]
Sure what do u need help with
4 0
3 years ago
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