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3 years ago

PLS ANSWER ASAP

Physics

1 answer:

lidiya [134]3 years ago

8 0

3,728.2271 miles per hour please mark brainliest

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A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68, and the

zvonat [6]

Explanation:

"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."

(556 x 0.68) = static friction of 378.08N. before movement occurs.

The forces (a) and (b) will not move it.

Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.

Forces (c) and (d) will move it, and accelerate it.

Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.

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3 years ago

Birds fly south for the winter.bats hibernate during winter.what are these examples of

Alecsey [184]

The animals need less food people it is winter time.

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2 years ago

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To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

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2 years ago

Match the vocabulary terms to their definitions.

Vilka [71]

See the attached picture:

8 0

3 years ago

Read 2 more answers

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

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

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago