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3 years ago

Match the measurement with the proper Sl unit Acceleration: Velocity Distance:

Physics

2 answers:

Ket [755]3 years ago

6 0

Answer:

Acceleration - ms^-2

Velocity- ms^-1

Distance-m

Xelga [282]3 years ago

3 0

Answer:

Acceleration:

$m/s^{2}$

Velocity:

$m/s$

Distance:

$m$

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Why are there temperature differences on the moon's surface even though there is no atmosphere present?

nika2105 [10]

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.

4 0

3 years ago

9) What will happen to the period of a pendulum if we change the rope of a pendulum with another which is four times with the in

Arturiano [62]

Double

Explanation:

Since the period T of a pendulum is given by

$T = 2\pi \sqrt{\dfrac{l}{g}}$

By increasing the length of the pendulum by 4, the period becomes

$T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T$

You can see that the period doubles when we increase the length by a factor of 4.

5 0

2 years ago

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and han

sleet_krkn [62]

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

4 0

3 years ago

Which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar

MrMuchimi

The answer to the question is A

7 0

3 years ago

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

andreev551 [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0

3 years ago