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MAXImum [283]
3 years ago
11

G A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after t t

Physics
1 answer:
bija089 [108]3 years ago
4 0

Complete question:

A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after 7s.

Answer:

The final velocity of the dragster is 245 m/s.

Explanation:

Given;

acceleration of the dragster, a = 35 m/s²

initial velocity of the dragster, u = 0

time of motion, t = 7 s

the final velocity of the dragster after 7s is given by;

v = u + at

where;

v is the final velocity

u is the initial velocity

v = 0 + (35 x 7)

v = 245 m/s

Therefore, the final velocity of the dragster is 245 m/s.

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What is the displacement from a starting position of (14.0, 3.0) m to a final position of (−3.0, −4.0) m?
dem82 [27]

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3 years ago
A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w
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Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85  = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x  5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

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4 0
3 years ago
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

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now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
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