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4 years ago

When a solo eclipse occurs. It is seen in

Physics

1 answer:

mr_godi [17]4 years ago

4 0

The sky? That is the only reasonable answer I can come up with using the information you provided.
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A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the fo

Elina [12.6K]

Answer:

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

3.9t = 22.3

t = 5.72 s

5 0

3 years ago

Read 2 more answers

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

4 years ago

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
$F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

For more information on this visit

brainly.com/question/21811998

5 0

3 years ago

A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?

belka [17]

Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

The time period of the ball, t = 8 s

The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

v = d / t m/s

Substituting the given values in the above equation,

v = 24 / 8

= 3 m/s

Hence, the speed of the ball was, v = 3 m/s

8 0

3 years ago

What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp

Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

$Q=mC\Delta T$

where:

m is the mass of the substance

C is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

$\Delta T = 15^{\circ}C$ (change in temperature)

Substituting, we find:

$Q=(25)(0.39)(15)=146 J$

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0

3 years ago