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3 years ago

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Physics

1 answer:

Mila [183]3 years ago

5 0

Answer: When you break on your bike and when you rub your hands together to get warm.

Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop. The rubbing hands is making friction. Which makes you get warm.

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What characteristic do the metals iron, nickel, and cobalt share?

Rzqust [24]

They are all conductors.

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3 years ago

Read 2 more answers

In the space around a permanent magnet, where is the magnetic field the strongest?

BlackZzzverrR [31]

near the north and south poles of the magnet

Explanation:

Magnetic fields around a permanent magnet is strongest near the north and south poles of the magnet.

Magnetic fields are the region of space around a magnet where magnetic effects are felt.

This is as a result of a force field that surrounds the magnet.
Magnetic fields are strongest within the magnet.
Also, externally, they are strongest at the poles of a magnet.
Around the poles, magnetic lines of force leaves and enters a magnet.

learn more;

Electromagnet brainly.com/question/2191993

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6 0

3 years ago

Identical twins, each with mass 61.0 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twi

Anna007 [38]

Answer:

Immediately after throwing the backpack away, twin A would be moving away from twin B at approximately $0.630\; \rm m \cdot s^{-1}$ .

Initially, twin B would not immediately be moving. However, after the backpack hits her, she would move away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ if she held onto the backpack.

Explanation:

Consider this scenario in three steps:

Step one: twin A is carrying the backpack.
Step two: twin A throws the backpack away; the backpack is en route to twin B;
Step three: twin B starts to move after the backpack hits her.

Since all external forces are ignored, momentum should be conserved when changing from step one to step two, and from step two to step three.

In step one, neither twin A nor the backpack is moving. Their initial momentum would be zero. That is:

$p(\text{twin A, step one}) = 0$ .
$p(\text{backpack, step one}) = 0$ .

Therefore:

$p(\text{backpack, step one}) +p(\text{twin A, step one}) = 0$ .

In step two, the backpack is moving towards twin B at $3.20\; \rm m \cdot s^{-1}$ . Since the mass of the backpack is $12.0\; \rm kg$ , its momentum at that point would be:

$\begin{aligned}p(\text{backpack, step two}) &= m \cdot v \\ &= 12.0\;\rm kg \times 3.20\; \rm m \cdot s^{-1} = 38.4\; \rm kg\cdot m \cdot s^{-1} \end{aligned}$ .

Momentum is conserved when twin A throws the backpack away. Hence:

$\begin{aligned}&p(\text{backpack, step two}) +p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one})\end{aligned}$ .

Therefore:

$p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one}) - p(\text{backpack, step two}) \\ &= -38.4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

The mass of twin A (without the backpack) is $61.0\; \rm kg$ . Therefore, her velocity in step two would be:

$\begin{aligned} v(\text{twin A, step two}) &= \frac{p}{m} \\ &= \frac{-38.4\; \rm kg \cdot m \cdot s^{-1}}{61.0\; \rm kg} \approx -0.630\; \rm m \cdot s^{-1}\end{aligned}$ .

Note that while the velocity of the backpack is assumed to be greater than zero, the velocity of twin A here is less than zero. Since the backpack is moving towards twin B, it can be concluded that twin A is moving in the opposite direction away from twin B.

<h3>From step two to step three</h3>

In step two:

$p(\text{twin B, step two}) = 0$ since twin B is not yet moving.
$p(\text{backpack, step two}) = 38.4\; \rm kg \cdot m\cdot s^{-1}$ from previous calculations.

Assume that twin B holds onto the incoming backpack. Thus, the velocity of the backpack and twin B in step three will be the same. Let $v(\text{twin B and backpack, step three})$ denote that velocity.

In step three, the sum of the momentum of twin B and the backpack would thus be:

$\begin{aligned}& m(\text{twin B}) \cdot v(\text{twin B and backpack, step three}) \\ &+ m(\text{backpack}) \cdot v(\text{twin B and backpack, step three})\end{aligned}$ .

Simplify to obtain:

$(m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})$ .

Momentum is conserved when twin B receives the backpack. Therefore:

$\begin{aligned}& (m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})\\ =&p(\text{twin B, step two}) +p(\text{backpack, step two})\\ =& 38.4\; \rm kg \cdot m\cdot s^{-1} \end{aligned}$ .

Therefore:

$\begin{aligned}& v(\text{twin B and backpack, step three})\\ =&\frac{p(\text{twin B, step two}) +p(\text{backpack, step two})}{m(\text{twin B}) + m(\text{backpack})}\\ =& \frac{38.4\; \rm kg \cdot m\cdot s^{-1}}{61.0\; \rm kg + 12.0\; \rm kg} \approx 0.526\;\rm m\cdot s^{-1} \rm \end{aligned}$ .

In other words, if twin B holds onto the backpack, then (after doing so) she would be moving away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ .

6 0

3 years ago

A 34kg crate resting on a horizontal surface is pushed as shown. If the coefficient of static friction s between the surfaces is

ioda

Answer:

D (not so sure, look in the explanation and judge yourself)

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards

Minimum force required = 0.25 x 34 x 9.81

= 83.385 N

So technically the force need to be greater than 83.385 so the answer should be d, not really sure if it should be b though really depends on the acceleration due to gravity.

7 0

3 years ago

An amusement park designer wants his swing to have a centripetal acceleration of 24 m/s2. If the velocity of the swing is 11 m/s

viva [34]

Answer:

radius is C: 5.0 m

Explanation:

formular for centripetal acceleration:

$a = \frac{ {V}^{2} }{r}$