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madreJ [45]
3 years ago
9

How can density be determined in a lab of a rectangular solid?

Physics
1 answer:
True [87]3 years ago
6 0
We will first record its mass and then its volume by measuring its dimensions
then divide mass by volume and will get density of regular solid
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according to newton's first law of motion what is the reason for a ball throwing up in the airfall back to earth
skelet666 [1.2K]
The force of earth's gravitational field is always directed downwards (towards the center of the earth. When the ball is thrown up, it is going against the earth's gravitational field and so, the earth's gravitational force pulls it back down, accelerating it downwards.
5 0
3 years ago
Does changing position of charges change the magnitude
hodyreva [135]

Answer:

No, they will not change.

Explanation:

7 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
The base of a hexagonal prism has an apothem measuring 10 inches and an area of 346.41 square inches. the prism is 12 inches tal
tigry1 [53]

The approximate lateral area of the prism is determined as 831 square inches.

<h3>What is lateral area of the hexagonal prism?</h3>

The lateral area of the hexagonal prism is calculated as follows;

LA = PH

where;

  • P is perimeter of the prism
  • H is height

A = ¹/₂Pa

where;

  • a is apothem = 10 inches
  • A is base area = 346.41 in²

346.41 = ¹/₂(10)P

346.41 = 5P

P = 346.41/5

P = 69.282 inches

LA = PH

LA = 69.282 x 12

LA = 831.38 in²

Learn more about lateral area of prism here: brainly.com/question/296674

#SPJ4

8 0
2 years ago
Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons a
dsp73

Answer:

E=576.5V/m

Explanation:

From the question we are told that:

Length l=56.0cm=0.56m

Distance apart d=7.0mm=0.007m

Electron Transferred n=10^{-10}

Therefore

Total Charge

Since Charge on each electron is

e=1.602*10^{-19}

Therefore

T=1.602*10^{-19} *10^{10}

T=1.602*10^{-9}

Generally the equation for Charge density is mathematically given by

\rho=T/A

Where

Area

A=0.56*0.56

A=0.3136

Therefore

\rho=1.602*10^{-9}/0.3136

\rho=5.10*10^{-9}

Generally the equation for Electric Field in the capacitor is mathematically given by

E=\frac{\rho}{e_0}

E=\frac{5.10*10^{-9}}{8.85x10{-12}}

E=576.5V/m

8 0
3 years ago
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