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3 years ago

“What are three ways that a cat pushing on a cat-flap door can change the amount of torque applied to the door?”

1 answer:

7 0

It depends on where the cat is applying the force and how much the force is .. after all where and at what is distance from the axis of rotation of the door

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Gravity affects atmospheric pressure true or false

Burka [1]

Answer:

false

Explanation:

5 0

3 years ago

At what angle two forces P + Q and (P - Q) act so that their resultant is :

stiv31 [10]

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

8 0

2 years ago

Read 2 more answers

"An object at rest starts accelerating. If it is going 120 m/s after traveling 202 meters, how quickly did it speed up?"

Marizza181 [45]

Answer:

this is just a guess bc i only looked at it for 5 seconds but i think 150 m/s

4 0

3 years ago

How much force is required to move a sled 5 meters if a person uses 60 j of work? 300N 12N 65N 30N

maksim [4K]

The answer is b 12N because
We know that
W = F × d × c o s(θ)
assuming theta=0 we then solve and have
F=W/d
substitute known values to get:
F=60J/5m=12N

4 0

3 years ago

Calculate the work done by the cyclist when his power output is 200 W for 1800 seconds.

Kobotan [32]

Answer:

3.6 × 10^5 J

Explanation:

Work Done = Energy

Energy = Power × time

Energy = 200 × 1800

= 360000 J

W =

$3.6 \times 10 {}^{5} j$

3 0

3 years ago