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Anika [276]
2 years ago
12

A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

is driving, the drawbridge is raised to an incline of 30°. What is the car's new acceleration on this incline, ignoring the force due to air resistance? (recall that g=9.8 m/s^2)
Physics
1 answer:
Viktor [21]2 years ago
7 0

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N

So now we can apply Newton's second law to find the acceleration of the car:

F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2

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Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

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2 years ago
In any electromagnetic wave,
KATRIN_1 [288]
C.half the energy is carried by the electric field and half is carried by the magnetic field.
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Write the first equation of motion. Under what condition(s) is this equation valid?​
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Explanation:

The first equation of motion in kinematics is given by :

v=u+at .....(1)

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Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be
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Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
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a flashlight operates with a current of 3.0 a and a power of 4.5 W. what is the voltage of the flashlights battery?
Dafna11 [192]

Answer:

V=1.5V

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