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2 years ago

Samples of different materials, A and B, have the same mass, but the sample

Physics

1 answer:

madam [21]2 years ago

4 0

Answer:

A.) The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

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What are the Properties of Transverse Waves and Longitudinal wave

jeyben [28]

Answer:

Key terms

TermMeaningTransverse waveOscillations where particles are displaced perpendicular to the wave direction.Longitudinal waveOscillations where particles are displaced parallel to the wave direction

7 0

3 years ago

A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0

3 years ago

Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th

Vladimir79 [104]

Answer:

$C = 1.77 \times 10^{-4} F$

Explanation:

As we know that in AC circuit we have

$V = i x_c$

here we have

V = 59 V

i = 5.05 A

so we will have

$x_c = \frac{59}{5.05}$

$x_c = 11.68 ohm$

also we know that

$x_c = \frac{1}{\omega C}$

here we will have

$11.68 = \frac{1}{(2\pi 77)C}$

$C = 1.77 \times 10^{-4} F$

7 0

3 years ago

Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.

LUCKY_DIMON [66]

Explanation:

Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.

The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.

The relationship between Gibbs free energy change and reaction quotient of the reaction is:

$\Delta G=\Delta G^o+RT ln Q$ ......(1)

where,

$\Delta G$ = Gibbs free energy change

$\Delta G^o$ = Standard Gibbs free energy change

R = Gas constant

T = Temperature

At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:

$\Delta G^o=-RT ln Q$ ...(2)

4 0

3 years ago

The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in

vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

m is mass of the rocket
v is the final velocity of the rocket
u is the initial velocity of the rocket
t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0

2 years ago