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3 years ago

What would happen if both the radius and mass of a planet were cut in half

Physics

1 answer:

Shtirlitz [24]3 years ago

5 0

Explanation:

According to formula

g = GM/R^2

when mass is halved the value of g becomes half but when radius is halved the value of g increases 4 times.

As a result of both value of g becomes twice.

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When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be

Wittaler [7]

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

$V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]$

We now know that density is defined as the relationship between mass and volume.

$Ro = m/V_{d}$

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

$Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]$

7 0

3 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

power drawn by the car, $P=8.8\ kW$

Now the Current delivered to the motor:

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

Distance travelled before battery is out of juice:

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

Now the time of operation of the car:

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

Now the distance travelled:

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

3 years ago

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

butalik [34]

Answer: a) 2,508

Explanation:

6 0

3 years ago

Can someone give me 5 differences for each velocity and speed. FOR EACH...

VARVARA [1.3K]

Answer:

132-123+ere

Explanation:

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5 0

3 years ago

Read 2 more answers

A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from

Bezzdna [24]

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

$B = \frac{\mu_o I}{2\pi R}$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

$B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T$

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.

7 0

3 years ago