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oksano4ka [1.4K]
2 years ago
12

A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu

me a
constant change in velocity.
A. 6000 N
B. 12,000 N
C. 7000 N
D. 18,000 N
h​
Physics
1 answer:
serg [7]2 years ago
5 0

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

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Read the scenario below and answer the question that follows. Susan was in her algebra class preparing to take a test. Her instr
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Explanation:

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3 years ago
50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0
2 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
Delvig [45]

Answer:

E=\dfrac{\lambda }{2\pi \varepsilon _or}

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}

E_w=\dfrac{\lambda }{2\pi \varepsilon _or}

Now the electric filed due to hollow metal cylinder

E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}

E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}

Now  by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

E=\dfrac{\lambda }{2\pi \varepsilon _or}

5 0
3 years ago
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
ella [17]

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

4 0
2 years ago
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