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2 years ago

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Two basketball teams are resting during halftime. While they are resting, a truck driver asks them for help pushing his broken d

own truck to the top of a nearby hill so he can coast to the mechanic's shop. One team agrees to help, but the other refuses to help. The team that goes to push the truck is unable to get the truck to the top of the hill by themselves.

1 answer:

viva [34]2 years ago

7 0

Answer:

It is a superordinate goal because both teams could have helped with the task.

Explanation:

If both teams pushed then they could have made it happened

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A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat

Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

$v=u+at$ ----------1

$s=ut+\dfrac{1}{2}at^2$ ------2

From equation 1 and 2

$s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )$

So we can say that

$s= \left(\dfrac{v+u}{2}\right)t$

Given that the velocity of car at final condition will be zero (v=0)

$s= \left(\dfrac{0+u}{2}\right)t$

$s= \left(\dfrac{u}{2}\right)t$

From the above equation we can find the initial velocity of car without finding the acceleration

$20= \dfrac{u}{2}\times 4$

u = 10 m/s

7 0

2 years ago

The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p

zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

Explanation:

Given that,

Average pressure $P=1.01\times10^{5}\ Pa$

Radius of earth $R_{E}=6.38\times10^{6}\ m$

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

$P=\dfrac{F}{A}$

$F=PA$

$F=P\times 4\pi\times R_{E}^2$

Where, P = pressure

A = area

Put the value into the formula

$F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2$

$F=516.6\times10^{17}\ N$

Hence, The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

8 0

2 years ago

Read 2 more answers

A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?

yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0

2 years ago

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

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2 years ago

The diagram below depicts all the forces acting on an object. Use both vector resolution and vector

maksim [4K]

Answer:

A

Explanation:

i just took the test

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2 years ago