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arsen [322]
2 years ago
7

According to gay-lussac’s law: select one:

Physics
2 answers:
DaniilM [7]2 years ago
8 0
Gay-Lussac's Law states
P1 / T1 = P2 / T2
So the answer is b

svlad2 [7]2 years ago
4 0

Answer

D

Explanation:

Just did the assignment.

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A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
cricket20 [7]

Answer:

<em>The period of the motion will still be equal to T.</em>

<em></em>

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

T = 2\pi \sqrt{\frac{M}{k} }

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>

5 0
2 years ago
Answer pls will give 20 points per person
LuckyWell [14K]

Answer: question 1 is b I believe

Explanation:

every action has an opposite reaction

7 0
3 years ago
Read 2 more answers
A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3
PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0
2 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

F_f = \mu m g

So the acceleration of car due to friction is given as

F_{net} = - \mu mg

a = \frac{F_{net}}{m}

a = -\mu g

now it is given that

\mu = 0.868

g = 9.81 m/s^2

so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0
3 years ago
Calculate the speed of the car at each checkpoint by
lana66690 [7]

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

3 0
2 years ago
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