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CaHeK987 [17]
3 years ago
15

Which is a standard unit of measurement​

Physics
1 answer:
ruslelena [56]3 years ago
6 0
There are some more like Time - second (s)
Amount of substance - mole (mole)
Electric current - ampere (A)
Temperature - kelvin (K)
Luminous intensity - candela (cd)

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Please help me , thank you
mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

5 0
3 years ago
Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k
NikAS [45]

Answer:

Fr=12544N

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation Fr=usmg, where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

Fr=0.8*1600kg*9.8\frac{m}{s^{2}}

Fr=12544N

5 0
3 years ago
Your moods is approved when you have extra stress hormones in your blood
jenyasd209 [6]
Is this a true or false question?
5 0
4 years ago
A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force
alukav5142 [94]

Answer:

497.6 N

Explanation:

From the question,

The net force on the skydiver = weight of the skydiver- the resistive force of air

F' = W-F...................... Equation 1

Where W = weight of the skydiver, F = resistive force of air.

But,

W = mg................ Equation 2

Where m = mass of the skydiver, g = acceleration due to gravity.

Substitute equation 2 into equation 1

F' = mg-F............ Equation 3

Given: m = 87 kg, F = 355 N, g = 9.8 m/s²

Substitute these values into equation 3

F' = 87(9.8)-355

F' = 852.6-355

F' = 487.6 N

6 0
3 years ago
Read 2 more answers
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e
Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0
3 years ago
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