A mud blob, 0.350 kg, is thrown at a wall at 10 m/s. The blob sticks to the wall. What is the change in momentum of the blob?

WILL GIVE BRAINLIEST IF CORRECT!

Norma-Jean [14]

Answer:

10000N

Explanation:

Given parameters:

Mass of the car = 1000kg

Acceleration = 3m/s²

g = 10m/s²

Unknown:

Weight of the car = ?

Solution:

To solve this problem we must understand that weight is the vertical gravitational force that acts on a body.

Weight = mass x acceleration due to gravity

So;

Weight = 1000 x 10 = 10000N

5 0

3 years ago

The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass

Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0

3 years ago

What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,

WITCHER [35]

The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force. This is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k, is a very large number

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

<u>Explanation: </u>

We know that the universal formula for the estimation of gravitational force is,

$F=G \times\left(\frac{m_{1} \times m_{2}}{r^{2}}\right)$

Where,

F, Gravitational force, $m_{1} \text { and } m_{2}$ - objects masses r- The relative distance between the two objects

G, Gravitational Constant $6.67 \times 10^{-11} \frac{N m^{2}}{k g^{2}}$

And, the electrostatic force between two scalar charges according to the Coulomb’s law is,

$F=k \times\left(\frac{q_{1} \times q_{2}}{r^{2}}\right)$

F, Electrostatic force between two charges, $q_{1} \text { and } q_{2}$ - Two scalar charges, r - The relative distance between two scalar charges, k- Coulomb’s constant $8.987 \times 10^{9} \frac{N m^{2}}{c^{2}}$

Now, on comparing the values of G and k, we can easily evaluate that eventually, the gravitational force will be lesser than the coulomb’s force.

Besides this, we can also judge this fact through various examples such as, a balloon rubbed with a cloth, easily sticks to the wall for some time. Opposing the gravitational force of the Earth which is not the case with the normal balloon.

It drops without having the electrostatic force between the wall and the balloon. This shows that the gravitational force draws lesser impact on objects as compared to the electrostatic force.

7 0

3 years ago

Read 2 more answers

Please help me, thank you

Mnenie [13.5K]

Answer:

a

Explanation:

7 0

3 years ago

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

3 0

4 years ago