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2 years ago

10

Why are experiments often performed in laboratories?

1 answer:

Marrrta [24]2 years ago

6 0

Answer:

B

Explanation:

Laboratories are usually controlled: however, in public places, there are way too many factors that could change an experiment

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What is the speed of a truck that travels 10 i’m in 10 minutes ?

fiasKO [112]

10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.

Kph is very common for vehicles:

10 km/10 min (60 min/hr) = 60 kph

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3 years ago

PLEASE HEELP!!!

Ann [662]

Answer:

The "pressure" of the electricity is electric potential. Electric potential is the amount of energy available to push each unit of charge through an electric circuit. The unit of electric potential is the volt. ... A volt is the force needed to move one amp through a conductor that has 1 ohm of resistance

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2 years ago

A 26-kg sled is on a snow-covered slope. The coeﬃcients of friction between the sled’s runners and the snow are µs = 0.096 and µ

sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction $\mu _s=0.096$

coefficient of kinetic friction $\mu _k=0.072$

In order to move sled from rest we need to provide a force greater than static friction which is given by

$f_s=\mu mg=0.096\times 26\times 9.8=24.46 N$

After Moving Sled kinetic friction comes in to play which is less than static friction

$f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N$

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

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2 years ago

Let x define the position of an object such that x =

timama [110]

See the attached picture for answers

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2 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

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3 years ago