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3 years ago

A 0.6kg ball accelerated at 55 m/s 2 . What force was applied?

Physics

1 answer:

adell [148]3 years ago

7 0

Answer: 33N

Explanation:

Newton's second Law of motion F = ma

F = (0.6)(55) = 33N

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How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

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H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

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7 months ago

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2 years ago

Read 2 more answers

Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat

aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

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Divide equation (1) by equation (2) we get

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5 0

2 years ago

Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c

dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

$\omega = 2\pi\frac{24}{20}$

$\omega = 2.4\pi rad/s$

Now we know that the tangential speed of the chain is given as

$v = r \omega$

so we have

$v = (3 cm)(2.4\pi)$

$v = 7.2 \pi cm/s$

$v = 432\pi cm/min$

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

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3 years ago

A 0.6kg ball accelerated at 55 m/s​ 2​ . What force was applied?

A 0.6kg ball accelerated at 55 m/s 2 . What force was applied?