Which of the following is an example of kinetic energy?

If the mass of a moving object decreases from 100 grams to 25 grams, what happens to its momentum

cluponka [151]

The momentum will become 1/4 as much.
0.100 v === 0.25 v

5 0

3 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$

The force on the particle $q_1$ is $-180.38\ \text{N}$ .

8 0

3 years ago

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts

Helga [31]

Answer:

μ = 0.03

Explanation:

In order for the trunk not to slide the frictional force between the turntable and the trunk must be equal to the unbalanced force applied on the trunk by the motion of the turntable. Therefore,

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma (Newton's second law of motion)

Frictional Force = μN = μW = μmg

Therefore,

ma = μmg

a = μg

μ = a/g

where,

μ = coefficient of static friction between the trunk and the turntable = ?

a = tangential acceleration of trunk = 0.3 m/s²

g = 9.8 m/s²

Therefore,

μ = (0.3 m/s²)/(9.8 m/s²)

μ = 0.03

6 0

3 years ago

What's the average speed of an object moving 3790 meters in 249 s.

NISA [10]

Heya!!

For calculate velocity, lets applicate formula

$\boxed{d = v * t}$

Δ Being Δ

d = Distance = 3790 m

t = Time = 249 s

v = Velocity = ?

⇒ Let's replace according the formula and clear "v":

$\boxed{v= 3790\ m / 249\ s}$

⇒ Resolving

$\boxed{ v = 15,22 \ m/s }$

Result:

The velocity is 15,22 meters per second.

Good Luck!!

4 0

2 years ago

Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. The takeoff velocity is achieved a

Maksim231197 [3]

Answer:

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2$

Acceleration of the frog is 2065.005 m/s²

6 0

3 years ago