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3 years ago

A car sets off from traffic lights. It reaches a speed

Physics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

1.5 m/s^2

Explanation:

a = velocity/time

27/18 = 1.5 m/s^2

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Excellent human jumpers can leap straight up to a

Shkiper50 [21]

The speed that the person needs to leave the ground will be 4.32m/s

From the question given,

Height = 95cm

Since the person leave the ground v = 0m/s

acceleration due to gravity g = 9.8m/s²

Using the equation of motion

v² = u² + 2as

a = -g (upward motion)

s = h (distance changes to height)

The equation will become:

0² = u² - 2gh

0² = u² - 2(9.8)(0.95)

u² = 18.62

u = √18.62

u = 4.32

Hence the speed that the person needs to leave the ground will be 4.32m/s

Learn more here: brainly.com/question/20352766

8 0

2 years ago

To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially

cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside a pipe between two points.

Thus

⇒ $A_{1}$ × $V_{1}$ = $A_{2}$ × $V_{2}$

⇒ $\frac{\pi }{4}$ × $d_{1} ^{2}$ × $V_{1}$ =

⇒ $d_{1} ^{2}$ × $V_{1}$ = $d_{2} ^{2}$ × $V_{2}$

⇒ $(3.2)^{2}$ × 110 = $(0.57)^{2}$ × $V_{2}$

⇒ $V_{2}$ = 3467 cm / sec

⇒ $V_{2}$ = 34.67 m / sec

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0

3 years ago

Machines

Ainat [17]

Explanation:

Hey there!!

Here, Given is,

Efficiency = 75%

VR = no. of pulleys = 5

Now,

$effeciency = (\frac{ma}{vr} ) \times 100\%$

$75\% = \frac{ma}{5} \times 100\%$

100% ma = 75%×5

$ma = \frac{75\% \times5}{100\%}$

$ma = \frac{375\%}{100\%}$

Therefore, MA is 3.75.

Hope it helps...

3 0

3 years ago

Read 2 more answers

A certain electric dipole consists of charges q and –q separated by distance d, oriented along the x-axis as shown in the figure

Anna71 [15]

Electric field due to a point charge is given by formula

$E = \frac{kq}{r^2 + \frac{d^2}{4}}$

now due to negative charge also the magnitude of electric field will be same

only difference is the direction of field due to negative charge is radially inwards

now we can say that net field due to these two charges is given as

$E = 2E_0cos\theta$

$E = 2\frac{kq}{r^2 + \frac{d^2}{4}}*\frac{d/2}{\sqrt{r^2 + \frac{d^2}{4}}}$

$E = \frac{kqd}{(r^2 + \frac{d^2}{4})^1.5}$

now it is given that distance r is very large than "d" so we can say

$E = \frac{kqd}{r^3}$

so above is the electric field due to dipole

7 0

3 years ago

A) What is the meaning of specific heat capacity ?

dem82 [27]

Explanation:

the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

3 0

3 years ago