An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the

natulia [17]

Answer:

The value is $A = 2.80 *10^{-4} \ m^2$

Explanation:

From the question we are told that

The operating temperature is $T = 2450 \ K$

The emissivity is $e = 0.350$

The power rating is $P = 200 \ W$

Generally the area is mathematically represented as

$A = \frac{P}{ e * \sigma * T^2}$

Where $\sigma$ is the Stefan Boltzmann constant with value

$\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4$

So

$A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}$

$A = 2.80 *10^{-4} \ m^2$

8 0

3 years ago

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

4 0

2 years ago

In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times

Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0

3 years ago

A simple pendulum of length of 1.37 m and mass of 6.66 kg is given an initial speed of 2.85 m/s at its equilibrium position. Det

yawa3891 [41]

Answer:

2.35 s

Explanation:

The period of a simple pendulum is expressed as;

T = 2π $\sqrt{\frac{L}{g} }$

Where

T is the period in seconds

L is the length in metres

g is acceleration due to gravity

T = 2π $\sqrt{\frac{1.37}{9.8}}$

T = 2.349 s

T = 2.35 s

8 0

3 years ago