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3 years ago

14

To produce change in motion, a force must be a(n)

1 answer:

Bas_tet [7]3 years ago

5 0

It'd be an unbalanced force

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What is the acceleration of a 10.5 kg mass pushed by a 50.5 n force

Lina20 [59]

Answer:4.8 m/s^2

Explanation:

mass=10.5kg

Force=50.5N

Acceleration =force ➗ mass

Acceleration =50.5 ➗ 10.5

Acceleration =4.8 m/s^2

8 0

3 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Katyanochek1 [597]

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

$I = MR_A^2$

Substituting $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$ (Radius of wheel)

$I = 1.9 * 0.33^2$

$= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

$\tau = FR_B - F_rR_A$

Substituting 135 N for $F_r$ (Force acting on sprocket), $\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

$\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

$\tau = \alpha I$

Now equating the two equation for torque

$F (0.0435) - 135 (0.33) = \alpha I$

Making F the subject

$F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

$F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

$= 1041.7N$

4 0

3 years ago

Read this /https://www.carbonbrief.org/polar-bears-and-climate-change-what-does-the-science-say

DerKrebs [107]

Melting ice would damage this polar bears habitat meaning the polar bear may decrease

5 0

3 years ago

What are the components of the "Earth Radiation Budget?"

Alex777 [14]

Answer:

The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.

Explanation:

I hope this helps also I hope you have a great day and a new year.

4 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago