If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?

Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ

Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

$F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}$

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

6 0

3 years ago

Point masses m1 m2 are placed at opposite ends

tiny-mole [99]

(a) $x = \frac{m_2L}{m_1+m_2}$

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is $E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)$

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

$m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}$

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at $\frac{m_2L}{m_1 + m_2}$

7 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just bef

stich3 [128]

Answer:

v = 7.67 m/s for L= 1m

Explanation:

Let's use the conservation of mechanical energy, at the highest point and the lowest point

Initial. Vertical ruler

Em₀ = mg h

Final. Just before touching the floor

$Em_{f}$ = K = ½ I w²

Em₀ = $Em_{f}$

m g h = ½ I w²

The moment of inertia of a ruler that turns on one end is

I = 1/3 m L²

Let's replace

m g h = ½ (1/3 m L²) w²2

g h = 1/6 L² w²

They ask for the speed of the end so the height h is equal to the length of the ruler

g L = 1/6 L² w²

The linear and angular variables are related

v = w r

w = v / r

In this case the point of interest a in strangers r = L

g L = 1/6 L² v² / L²

v = √ 6 g L

Let's calculate

Assume that the length of the meter is L = 1 m

v = √ (6 9.8 1)

v = 7.67 m/s

7 0

3 years ago

A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is relea

k0ka [10]

Answer:

$D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}$

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

$Y_{f}=Y_{o}-g*\frac{t^{2}}{2}$

$0=H-g*\frac{t^{2}}{2}$ solving for t:

$t=\sqrt{\frac{2H}{g} }$

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

$X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}$ Replacing values:

$D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}$

Simplifying:

$D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}$

4 0

3 years ago

A truck is traveling due north at 75 km/hr approaching a crossroad. On a perpendicular road a police car is traveling west towar

Gre4nikov [31]

Answer:

109.66km

Explanation:

Velocity is defined as the change in displacement of a body with respect to time.

Velocity = displacement/Time

Displacement = velocity * time

A truck is traveling due north at 75 km/hr approaching a crossroad and perpendicular road a police car is traveling west toward the intersection at 80 km/hr, then the resultant velocity is gotten using the pythagoras theorem since they are both perpendicular to each other.

v² = 75²+80²

v² = 5625+6400

v² = 12,025

v = √12,025

v = 109.66km/hr

If both vechicles reaches cross road in exactly one hour, the displacement of the truck with respect to the police car = 109.66km/hr * 1 hour

Displacement of the truck with respect to the police car = 109.66km

7 0

3 years ago