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3 years ago

5

You can enter compound units that are combinations of other units that are multiplied together.Torque can be calculated by multi

plying the force (N ) applied at 90∘ to the lever arm at a distance (m ) from the pivot point (point of rotation), the compound SI unit for the torque is N⋅m . If the force (at 90∘ to the lever arm) applied is 15 N and it is applied at 2.0 m from the pivot point (point of rotation), what is the torque on the lever?Express the torque using two significant figures and include the appropriate units.

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

3.0 x 10¹ Nm

Explanation:

Torque = F x r

Where F is force applied and r is perpendicular distance from pivot point . r

is also called lever arm

Here F = 15 N and r = 2.0 m

Torque

= 15 N X 2.0 m

= 3.0 10¹ Nm.

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Answer:

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Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

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3 years ago

A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

MrMuchimi

The bullet travels a horizontal distance of 276.5 m

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The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

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Substitute 0 m/s for $u_y$ , 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

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3 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

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Answer:

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From the question w are told that

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$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

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Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

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$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

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$6v_B -4 = 0$

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4 0

3 years ago