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3 years ago

6. If a vehicle needs 5s to complete 15m. Find the mean speed of it?

Physics

1 answer:

Karolina [17]3 years ago

4 0

Answer:

3m/s

Explanation:

Time=5s

Distance =15m

Speed=distance/time

Putting the values

Speed=15m/5s

Speed=3m/s is the answer

Hope it will help you. :)

You might be interested in

A climber pulls herself 8 meters upwards with a force of 150 Newtons. If it takes her 16 seconds to cover the 8 meters, how much

mr Goodwill [35]

Answer:

P = 75 W

Explanation:

given,

Distance, L = 8 m

Force,F = 150 N

Time, t = 16 s

Work by the climber

Work done = Force x displacement

W = F. L

W = 150 x 8

W = 1200 J

We know,

$Power =\dfrac{Work\ done}{time}$

$P =\dfrac{1200}{16}$

P = 75 W

Hence, Power climber is using to climb is equal to 75 W.

3 0

3 years ago

The acceleration due to gravity on the surface of Venus is 8.83 m/s2. An object with a mass of 5.23 kg has what weight on Venus?

Marat540 [252]

Weight = m times g = 5.23 times 8.83 = 46.18 N

6 0

3 years ago

A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the rem

scZoUnD [109]

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

$W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft$

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s$

One Saucer power is 500000000 lbft/s

7 0

3 years ago

The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :

Delicious77 [7]

The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

$x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )$

Data;

ω = π/2

k = 1.254N/m

Solving this

$\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4} )$

Let's calculate the maximum velocity.

$V_{m} =\frac{3.5\pi }{2}$

This is only possible when cos θ = -1

The maximum kinetic energy is

$K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2$

$w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m$

Using the value of spring constant, we can find the maximum potential energy.

$P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J$

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

#SPJ1

8 0

2 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

3 years ago