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Vladimir79 [104]

3 years ago

14

Calculate the velocity of the object in the graph during the following time interval 0-6 seconds. Also, during which time interv

al is the object accelerating and how can you tell?

1 answer:

cestrela7 [59]3 years ago

3 0

Its accelarating from 1-10 i know that because the line is going up. i hope i helped

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How does the sun's energy most directly influence precipitation in an area?

topjm [15]

The sun's energy influences climate in various ways. For example the latitudes at the equator receive more energy from the sun and therefore have warmer temperatures, On the other hand the sun's energy influences precipitation in a climate by driving the water cycle which determines precipitation.The sun is what makes the water cycle take place. That is the sun provides energy or heat to the earth; the heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds ( precipitation), that in turn give us rain

5 0

3 years ago

An airplane flies horizontally at constant speed in a straightline direction. Its state of motion is unchanging. In other words

Alex_Xolod [135]

Answer:

sum of all forces on the air plane must be ZERO

So both forces must be of same magnitude

Explanation:

As we know that airplane is moving with uniform speed is horizontal plane is a straight line

so the motion of the air plane is uniform without any acceleration

So we will have

$a = 0$

acceleration must be zero

now by Newton's law

$F_{net} = 0$

$F_1 + F_2 = 0$

so sum of all forces on the air plane must be ZERO

7 0

3 years ago

Please help! I haven't really been able to understand this topic :/

Anettt [7]

Ok but y I thought it was upside down tho...

4 0

3 years ago

Read 2 more answers

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

3 years ago

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

$\omega_i$ = Initial angular velocity = 78 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0

3 years ago