Please answer these! Will mark brainiest if correct!

Answer:

A) P = W / t = 6000 / 60 = 100 J/s == 100 watts

B) P = I V = .05 * 240 = 12 watts

C)  Bulb B) uses much less power

Answer:

300 J

Explanation:

GIVEN:

Force (F) = 10 N

Displacement (s) = 30 m

or displacement (d) = 30m

Work done (W) = ?

Therefore,

W = F×s

or W= F × d

= 10×30

= 300 J

Hope it helps

Comparing and analyzing the brightness of near and farther streetlight would support the student’s argument.

The brightness of nearer streetlight is more as compared to the brightness of the farther light because brightness decreases with the increase in distance.

So we can conclude that comparing and analyzing the brightness of near and farther streetlight would support the student’s argument.

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Answer:

True

Explanation:

If it is at right conditions and correct time (night time most likely) they will all be visible

The motorcycle's average velocity will be 6 m/sec. The ratio of the overall distance traveled to the total time taken by the body is the average velocity.

The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.

Total distance travelled by the  motorcycle;

d=20 m+10 m

d=30 m

The average velocity is found as;

[tex]\rm v_{avg}=\frac{d}{t} \\\\ \rm v_{avg}=\frac{30}{5} \\\\ v_{avg}= 6 \ m /sec[/tex]

Hence,the motorcycle’s average velocity will be 6 m/sec.

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Answer:

3+7=          7x3=                                21➗7=          

Explanation:

3+7= 10              7 8 9 10

7x3= 21                  7 14 21

21➗7=3        21 14 7

                 mark me brainliest                      please

Answer:

To be better at something

Explanation:

I have Accelerated at cooking

Answer:

Acceleration means to speed up

Explanation:

Answer:

a.one atom of fluorine and two atoms of oxygen

Explanation:

hope it help

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

[tex]\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9 = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C[/tex]

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

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Answer:

.Also we can change the frequency by decreasing wavelength , or increasing velocity

Explanation:

Answer:

23.24 m/s

Explanation:

2.1 s  *  (-7.6 m/s^2) = -  15.96 m/s   from original speed

39.2 - 15.96 = 23.24 m/s

78 percent nitrogen and 21 percent oxygen

Answer: 911.625

Explanation:

Answer:

911.625 is the answer of your question

Answer:

0.12m/s

Explanation:

v=λf

Given that, λ = 12cm = 0.12m

T = 1second

(A period T is the time required for one complete cycle of vibration to pass a given point)

frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz

therefore, v= 0.12 × 1 = 0.12m/s

Answer:

Most light sources can be classified into three main types: continuous, absorption, and emission. A hot, opaque object, like the filament in an incandescent light bulb, emits a continuous spectrum, having light of all wavelengths. A hot, dense gas is another example of an object that emits a continuous spectrum.

Explanation:

Answer:

A) 75m

Explanation:

distance = speed x time taken

Here given:

speed: 15 m/s

time: 5 second

So, distance covered:

15 x 5

75 m

Distance

Answer:

Her kinetic energy decreases and her potential energy increases.

Explanation:

Answer:

b

Explanation:

The formula for work done (W) is:

W = Force x Distance moved in the direction of force

This means that the object needs to move in the direction of the force, otherwise no work is done in that direction.

Hope this helps!

The total momentum before collision is

14.0 kg•m/s + (-9.00 kg•m/s) = 5.00 kg•m/s

If v is the velocity of ball B after colliding with ball A, the total momentum after collision is

(2.00 kg) (-5.00 m/s) + (3.00 kg) v

Momentum is conserved, so these quantities are equal. Solve for v :

5.00 kg•m/s = (2.00 kg) (-5.0 m/s) + (3.00 kg) v

5.00 kg•m/s = -10.0 kg•m/s + (3.00 kg) v

15.0 kg•m/s = (3.00 kg) v

v = (15.0 kg•m/s) / (3.00 kg) = 5.00 m/s

The direction of force refers to the path along which the force acts. The plane of application of force, on the other hand, is the surface or area where the force is applied

The direction of force refers to the path along which the force acts. It is the straight line in which the force is being applied. For example, if you push a box on the ground, the direction of force is the line of action of the force vector that points in the direction of the push.

The plane of application of force is the surface or area where the force is applied.

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Answer:

option C) the iris

Explanation:

2023

Answer:

4 m/s

Explanation:

v=u+ at  and  v^2=u^2+2as.

It is given that  v=0,t=5  s    and  s=10  m.

⇒0=u+5a  and  0=u^2+20a.

⇒a=−u/5  and  a=−u^2/20.

⇒u/5 = u^2/20⇒u=4  m/s.

⇒  The initial velocity is  4  m/s.

Answer:

Here is your answer

Explanation:

Given:

Given:Density of sea water (d) = 1020 kgm^{3}3

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.

Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.Hence, depth of sea is 1 km

Answer:

C = frequency × wavelength, where C is the speed of the photons which is the speed of light ~3×10^8

frequency = C/wavelength

f = 3×10^8/400×10^-9m

f = 3/4 × 10^15Hz

f = 0.75×10^15Hz

f = 7.5×10^14Hz

Answer:

Length of steel = 5.5 m

Increase in length = 0.0012 m

α of steel = 12 x 10-6/K

To find:

Temperature change needed ?

Calculation:

Let the required temperature change be ∆T. (in the photo)

So, temperature change of 18.1 Kelvin is required

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[tex]\huge{\color{red}{\fbox{\textsf{\textbf{Answer}}}}} [/tex]

1) 5 seconds

2) 75 meters

Explanation:

Refer to the attachment

The horizontal component of the velocity with which Ronnie kicks a playground ball is 12.3 m/s.

The horizontal component of the velocity with which Ronnie kicks a playground ball is calculated as follows;

Vx = Vcosθ

where;

Vx = 16 x cos(40)

Vx = 12.3 m/s

Thus, the horizontal component of the velocity with which Ronnie kicks a playground ball is 12.3 m/s.

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Answer:

D) 12.3

Explanation:

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