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3 years ago

How is physical change different from a chemical change?

Physics

1 answer:

Mrac [35]3 years ago

5 0

Hello there!

Physical changes do not change what a substance is, whereas a chemical change results in a new substance being formed, so they do change what a substance is.

I hope I could help you to solve this question. Have a great rest of your day! :)

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If one of the charges is doubled in magnitude while maintaining the same separation between the charges, what is the new magnitu

lilavasa [31]

<h2>Magnitude of force doubles.</h2>

Explanation:

Force between two charges is given by

$F=k\frac{q_1q_2}{r^2}$

where q₁ and q₂ are charges and r is the distance between them.

Here one charge is doubled with keeping all others same.

q₁ = 2 q₁

We have

$F_1=k\frac{q_1q_2}{r^2}\\\\F_2=k\frac{2q_1q_2}{r^2}\\\\\frac{F_1}{F_2}=\frac{k\frac{q_1q_2}{r^2}}{k\frac{2q_1q_2}{r^2}}\\\\\frac{F_1}{F_2}=\frac{1}{2}\\\\F_2=2F_1$

Magnitude of force doubles.

7 0

3 years ago

A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car

Furkat [3]

Answer:

Part a)

$a = 4.86 m/s^2$

Part b)

$d = 155.52 m$

Part c)

$v_f = 48.6 m/s$

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

$v_f = 87 mph = 38.88 m/s$

now we have

Part a)

acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{t}$

$a = \frac{38.88 - 0}{8}$

$a = 4.86 m/s^2$

Part b)

distance moved by car with uniform acceleration is given as

$d = \frac{v_f + v_i}{2} t$

$d = \frac{38.88 + 0}{2} 8$

$d = 155.52 m$

Part c)

As we know that the car start from rest

so final speed after t = 10 s

$v_f = v_i + at$

$v_f = 0 + (4.86)10$

$v_f = 48.6 m/s$

3 0

3 years ago

What is the main difference between the Schrödinger model and the Bohr atomic model?

netineya [11]

Answer:

Schrödinger believed that electrons could only exist in orbits, but Bohr stated that electrons could be found anywhere in the atom

Explanation: I got it right on test

4 0

3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$

To find Ny, we need to find the tension T.
For this, we can equate the net horizontal force.

$F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N$

Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

$N_y= (40*9.8)-(169.8*sin59)=246.4N$

Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

$N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N$

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

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2 years ago

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How does a force pumb works

ser-zykov [4K]

Answer:

A force pump can be used to raise water by a height of more than 10m, the maximum height allowed by atmospheric pressure using a common lift pump.

In a force pump, the upstroke of the piston draws water, through an inlet valve, into the cylinder. On the downstroke, the water is discharged, through an outlet valve, into the outlet pipe.

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3 years ago

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