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2 years ago

50 N

Physics

1 answer:

Mariana [72]2 years ago

5 0

The work is done by the force pushing the mower is determined as 500 J.

<h3>Work done by the force</h3>

The work done by the force is calculated as follows;

W = Fsinθ x d

where;

Fsinθ = Fx is the applied force parallel to the ground
d is the distance
θ is the angle of inclination

W = Fx(d)

W = 25 x 20

W = 500 J

Thus, the work is done by the force pushing the mower is determined as 500 J.

Learn more about work done here: brainly.com/question/8119756

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3 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

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Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

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If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

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$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

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1 year ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

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distance=0.124 m

Explanation:

$mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\ v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\ L=0.124m$

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