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3 years ago

Whicg of the following is not a force that would act on a bicycle as it moves down a street

1 answer:

3 0

<span>1. It will stay at rest due to inertia. 2. acceleration 3. The couch will move to the left with a force of 1 N. 5. 6 N 6. Friction pulls the skates' wheels westward. 7. velocity 8. its speed increases due to gravity 10. It will have a constant velocity until acted upon by outside forces. 11. 35 N 12. mass 13. sliding friction 14. The couch will move with a force of 6 N. 15. It is pushing up on the book. 16. gravity 18. There is less friction between the car tires and the ice than between the car tires and the road. 20. balanced. 21. Due to centripetal force, the water is pushing on the bottom of the bucket. 22. a high velocity and a large mass 23. Scientists in other countries use SI units when measuring speed</span>

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An elevator filled with passengers has a mass of 1.70×103kg. (a) The elevator accelerates upward from rest at a rate of 1.20m/s2

Anton [14]

Answer with Explanation:

We are given that mass of an elevator filled with passengers = $1.70\times 10^3 Kg$

a.a= $1.2 m/s^2$

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second law,the net force is given by

$F_net=\sum F=ma$

$F-mg=ma$

Substitute the values then we get

$F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N$

Hence, the tension in the cable supporting the elevator= $1.87\times 10^4 N$

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W= $1.7\times 10^3\times 9.8=1.67\times 10^4 N$

Hence, the tension in the cable during this time = $1.67\times 10^4 N$

c. Deceleration= $0.6 m/s^2$

t=3 s

We have to find the tension in the cable during deceleration.

$F-mg=-ma$

$F=mg-ma=m(g-a)$

$F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N$

Hence, the tension in the cable during deceleration= $1.56\times 10^4 N$ .

d.We have to find that the height of elevator moved above its original staring point .

We have to find the final velocity.

$y_1=ut_1+\frac{1}{2}a_1t_1^2$

$y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m$

Final velocity

$v_1=a_1t_1$

$v_1=(1.2)(1.5)=1.8 m/s$

$y_2=v_1t_1=(1.8)(8.5)=15.3 m$

The third distance $y_3$ is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

$y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m$

The total distance moved by elevator from its original position

$d=1.35+1.8+2.7=19.35 m$

Final velocity=0

5 0

3 years ago

A moving car of mass 1500 kg travels at a constant speed of 4.6 m/s. What is the car's average momentum?

Contact [7]

Answer:

Explanation:

c I think this is the answer

3 0

3 years ago

Which type of electromagnetic radiation is used to keep food warm?

n200080 [17]

Answer:

A. infrared

Explanation:

Heat lamps used to keep foods warm are infrared on the light spectrum

7 0

3 years ago

The output power of a solar panel is the rate of transfer of...?

gtnhenbr [62]

Power is the rate of generating, moving, or using energy.

5 0

3 years ago

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch

IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0

4 years ago