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3241004551 [841]
3 years ago
11

a sleepy atudent drops a calculator out of a window yhata 20.7 m off the geound. we can ignore air resistance. what js the veloc

ity of the calculator after falling for 1.8 s
Physics
1 answer:
frez [133]3 years ago
6 0

The acceleration of gravity is

9.8 m/s^2 down.

When an object falls out of a hand, its speed after 1.8s is

(9.8)x(1.8) = 17.6 m/s down.

It doesn't matter what it is, how much it weighs, or how high it was dropped from.

If it's more than 17.6 m/s, then this happened on a different, bigger planet.

If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.

You might be interested in
Why is it important to select a coordinate system when studying motion?
kvv77 [185]

Motion is detected when an object changes its position with respect to a reference point. Coordinate system is basically used to represent motion. A coordinate system uses numbers or coordinates which represent position of the reference points on a two-dimensional or three-dimensional space. The trajectory of a point or line can be studied on a coordinate system which describes various aspects of motion like velocity, acceleration, distance, displacement etc. Coordinate system is important because it helps to choose a starting point and the direction (which will be positive).

5 0
3 years ago
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
a 2500 kg car going 35 m/a hits a 4000kg truck that is sitting still. What is the velocity of the truck if the car transfers all
lianna [129]

Answer:

Velocity of truck is 21.875 m/s.

Explanation:

Given:

Mass of the car (m) = 2500 kg

Initial speed of the car (u) = 35 m/s

Mass of the truck (M) = 4000 kg

Initial speed of the truck (U) = 0 m/s (Rest)

As per question, the total momentum of the car gets transferred to the truck.

Therefore, the final momentum of the car is 0.

Momentum is given as the product of mass and velocity.

So, if momentum becomes zero means the velocity becomes 0.

Therefore, final velocity of the car (v) = 0 m/s

Let the final velocity of truck be 'V' m/s.

Now, during collision, the total momentum remains conserved.

Initial momentum  = Final momentum.

Initial momentum of car + Initial momentum of truck = Final momentum of car + Final momentum of truck.

⇒ mu+MU=mv+MV\\\\2500\times 35+4000\times 0=2500\times 0+4000\times V\\\\87500+0=0+4000V\\\\4000V=87500\\\\V=\frac{87500}{4000}\\\\V=21.875\ m/s

Therefore, the velocity of the truck is 21.875 m/s.

7 0
3 years ago
A driver of a car traveling at 14.6 m/s applies the brakes, causing a uniform deceleration of 1.2 m/s2. How long does it take th
vfiekz [6]

Recall that

v_f=v_0+at

It takes the car about 3.2 s to reduce its speed from 14.6 to 10.8 m/s, since

10.8\,\dfrac{\mathrm m}{\mathrm s}=14.6\,\dfrac{\mathrm m}{\mathrm s}+\left(-1.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=3.2\,\mathrm s

Next, recall that

\Delta x=x_f-x_0=\dfrac{v_f+v_0}2t

Then the car undergoes a displacement of about 41 m, since

\Delta x=\dfrac{10.8\,\frac{\mathrm m}{\mathrm s}+14.6\,\frac{\mathrm m}{\mathrm s}}2(3.2\,\mathrm s)\implies\Delta x=41\,\mathrm m

6 0
3 years ago
Which of the following force fields will deflect electron beams?
inysia [295]
I'm going to have to go with A Magnetic because it seems to be the only one that makes sense it can't B because electron and Electrical are the same thing its not going deflect it and C it might seem right but gravity has nothing to do with deflection so I pick Magnetic.

So your answer is A.
6 0
3 years ago
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