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2 years ago

Which forces are contact forces give one example of each

Physics

1 answer:

insens350 [35]2 years ago

8 0

Answer:

A contact force is any force that requires contact to occur. Contact forces are ubiquitous and are responsible for most visible interactions between macroscopic collections of matter. Pushing a car up a hill or kicking a ball across a room are some of the everyday examples where contact forces are at work.

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A train is moving in uniform velocity and acceleration. A girl inside the train tosses a coin. Where would the coin land?

Furkat [3]

On the floor because the force from the train moving and she wouldn't be able to catch it in time

7 0

2 years ago

Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by

Ray Of Light [21]

Answer:

2.47 m/s backwards

Explanation:

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂...................... Equation 1

Where m₁ and m₂ = mass of the first basketball player and second basket ball player respectively, u₁ and u₂ = initial velocity of the first basket player and the second basketball player respectively, v₁ and v₂ = The final velocity of the first basket ball player and second basket ball player respectively.

Making v₁ the subject of the equation,

v₁ = (m₁u₁ + m₂u₂ - m₂v₂)/m₁.......................... Equation 2.

Given: m₁ = 87.2 kg, m₂ = 102.0 kg, u₁ = 7.0 m/s, u₂ = -5.2 m/s, v₂ = 2.9 m/s

Note: u₂ is negative because it moves towards the first basket ball player.

Substitute into equation 2

v₁ = [87.2(7.0)+102(-5.2) - (102×2.9)]/87.2

v₁ = (610.4-530-295.8)/87.2

v₁ = -215.4/87.2

v₁ = -2.47 m/s.

Thus the velocity of the 87.2 kg player = 2.47 m/s backwards.

7 0

3 years ago

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

Now we can solve for $v$ . We find that

$v=20.8\,\dfrac{\mathrm m}{\mathrm s}$

In the case that the ratio of accelerations is actually

$\dfrac{a_2}{a_1}=1.66$

we would instead have

$\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)$

in which case we would get a velocity of

$v=24.4\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

How to get to the daley center from pink line?

garik1379 [7]

From Pink Line, you use public transit to get to Daley Center. It will take around 16 minutes. The Richard Daley Center is located in between the streets of Randolph and Washington, and <span>Dearborn and Clark. The Bongo Room is a famous landmark near the Daley Center.</span>

6 0

3 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

2 years ago

Which forces are contact forces give one example of each​

Which forces are contact forces give one example of each