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Dovator [93]
3 years ago
15

You are asked to design an experiment which is similar to Match Graph activity. If you move 3m back from the motion sensor in 2s

. What is your speed?
Physics
1 answer:
Art [367]3 years ago
5 0

Answer:

The speed is 1.5 m/s.

Explanation:

Given that,

Distance = 3 m

Time = 2 sec

We know that,

The position and time graph shows the speed of motion.

We need to calculate the speed

Using formula of speed

v=\dfrac{d}{t}

Where, d = distance

t = time

Put the value into the formula

v=\dfrac{3}{2}

v=1.5\ m/s

Hence, The speed is 1.5 m/s.

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A car accelerates from rest to 19 m/s in 6 seconds. What is the acceleration of the car?
choli [55]

Answer:

the car is moving so that how it gos so fast

Explanation:

7 0
3 years ago
A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
pashok25 [27]

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

5 0
3 years ago
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
Read 2 more answers
A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef
Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.


Initial horizontal velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s 

15.5m / 10.3923m/s = 1.49s 

So the rock was in the air for 1.49 seconds. </span>

<span>

Now that we know that, we can use the following kinematics equation: 

d = v i * t + 1/2 * a * t^2 

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>

<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s 

So: 

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2 
d = 8.94 + -10.89</span>

d = -1.95<span>

<span>This means that the initial y position is 1.95 m higher than where the rock lands. </span></span>

5 0
3 years ago
If your acceleration speed is 15 m/s and you weigh 155 lbs, what is your gravitational force?​
sp2606 [1]

Answer:

Gravitational force, F = 1054.65 N

Explanation:

Given,

The accelerating speed, a = 15 m/s²

The mass of your body, m = 155 lbs

                                           = 70.31 Kg

The gravitational force acting in a body is given by the relation

                                F = m x g

Where g is the acceleration due to gravity of the in which the velocity of the body changes its speed at a constant rate.

                   ∴            a = g

Substituting the values in the above equation

                                  F = 70.31 x 15

                                     = 1054.65 N

Hence, the gravitational force acting on you, F = 1054.65 N

6 0
3 years ago
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