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skad [1K]
3 years ago
8

When adding vectors in the x and y direction, what must you do to find the magnitude and direction of the resultant vector?

Physics
1 answer:
RSB [31]3 years ago
6 0

Apply the equation. to find the magnitude, which is 1.4.

Apply the equation theta = tan–1(y/x) to find the angle: tan–1(1.0/–1.0) = –45 degrees. However, note that the angle must really be between 90 degrees and 180 degrees because the first vector component is negative and the second is positive.

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A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
rosijanka [135]

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

5 0
3 years ago
A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

             Initial momentum = Final momentum

                                       Pi = Pf

                  M*vs_e = m*vc_e + 4m*vm_e

Where,

                  M = m + 4m = 5m

                  vc_e = Velocity of command relative to earth

                  vm_e = Velocity of motor relative to earth  

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

                  vm_e =  v_c/m + vc_e        

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

                  5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

                  5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

                  5m*vs_e -  4m*(v_c/m) = 5m*vc_e

                   vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

                  vc_e = 3760 - 0.8*(90)

                  vc_e = 3,688 km/h

5 0
3 years ago
Read 2 more answers
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the
alexandr402 [8]

Answer:

r=6.05km/hr

z=59.1 degree to the horizontal

Explanation:

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place

can be solved using pythagoras theorem

r^2=o^2+a^2

r^2=5.2^2+3.1^2

r^2=36.65

r=6.1km/hr is te birds velocity relative to the ground

tanz=5.2/3.1

z=tan^-1(5,2/3.1)

z=59.1 degree to the horizontal

7 0
3 years ago
Read 2 more answers
If your average speed is 3 m/s, how far have you traveled in 1 second, 2 second, 3 seconds?
vichka [17]

Answer:

in 1 second 3m, in 2 seconds 6m, in 3 seconds 9m.

Explanation:

distance=speed × time

4 0
3 years ago
Read 2 more answers
a 2,000 pound car is driving at 60 miles/hour along a straight, level road. what is the net force acting on the car?
SVETLANKA909090 [29]

Answer:

0

Explanation:

According to Newton's second law, the net force is equal to the mass times the acceleration.  Since the car is not accelerating, the net force is 0.

5 0
3 years ago
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