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2 years ago

6

A thug pushes some 100 kg punk with 300 N of force. How much is the punk accelerated?

1 answer:

astra-53 [7]2 years ago

5 0

Answer:

Acceleration generate by punk = 3 m/s²

Explanation:

Given:

Weight of punk = 100 Kg

Force applied on punk = 300 N

Find:

Acceleration generate by punk = ?

Computation:

Acceleration = Force / Mass

Acceleration generate by punk = Force applied on punk / Weight of punk

Acceleration generate by punk = 300 N / 100 Kg

Acceleration generate by punk = 3 m/s²

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All known frequencies of the visible spectrum are..

crimeas [40]

Explanation:

$b. \: light$

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3 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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3 years ago

Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.

vladimir1956 [14]

By Snell's law:

η = sini / sinr. i = 25, η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.

r = sin⁻¹(0.3177)

r ≈ 18.52°

Option A.

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3 years ago

Converting 15 miles to kilometer

Dmitry [639]

15 miles to kilometers would be: 24.14 kilometers

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dmitriy555 [2]

Answer:

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Explanation:

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3 years ago