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2 years ago

A 23.7 kg kid slides down a

Physics

1 answer:

anyanavicka [17]2 years ago

6 0

Answer:

$\text { The acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

Mass of the kid 23.7 kg.

$\text { The kid is accelerating down at an angle is } 47.2^{\circ} .$

$^{\prime \prime} \mathrm{g}^{\prime \prime} \text { acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}$

We need to find the acceleration of the kid,

We know that, Parallel force acted on the kid at an angle is

F = m × g × sinθ (F = ma)

m × a = m × g × sinθ

Now, substitute the given values in the above formula to find acceleration of the kid,

$23.7 \times a=23.7 \times 9.8 \times \sin 47.2^{\circ}$

23.7 × a = 232.26 × 0.733

23.7 × a = 170.24

$a=\frac{170.24}{23.7}$

$a=7.18 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}$

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Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average

ryzh [129]

Answer:

9517.2 lbm

Explanation:

Electricity consumption = 14000 kWh/year

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3 years ago

You are on a 4 m high ladder and throw a ball upwards at 12 m/s. It lands on the ground below the ladder.

Gelneren [198K]

a) 2.75 s

The vertical position of the ball at time t is given by the equation

$y= h+ut-\frac{1}{2}gt^2$

where

h = 4 m is the initial height of the ball

u = 12 m/s is the initial velocity of the ball (upward)

g = 9.8 m/s^2 is the acceleration of gravity (downward)

We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

$0 = 4 + 12t - 4.9 t^2$

This is a second-order equation. By solving it for t, we find:

t = -0.30 s

t = 2.75 s

The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.

b) -15.0 m/s (downward)

The final velocity of the ball can be calculated by using the equation:

$v=u-gt$

where

u = 12 m/s is the initial (upward) velocity

g = 9.8 m/s^2 is the acceleration of gravity (downward)

t is the time

By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:

$v=12 -(9.8)(2.75)=-15.0 m/s$

And the negative sign means the direction is downward.

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2 years ago

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s

AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight $W = 150$ N

Length of rope $r = 4.1$ m

Initial speed of ball $v = 5.5 \frac{m}{s}$

For finding the tension in the rope,

First find the mass of rod,

$mg = 150$ ( $g = 9.8 \frac{m}{s^{2} }$ )

$m = \frac{150}{9.8}$

$m = 15.3$ kg

Tension in the rope is,

$T = mg + \frac{mv^{2} }{r}$

$T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}$

$T = 262.88$ N

Therefore, the tension in the rope is 262.88 N

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3 years ago

How many neutrons are needed to initiate the fission reaction shown?

nalin [4]

Answer: One neutron

Explanation:

one neutron 1/0n

Sum up the mass numbers on the right 99 + 135 + 2 = 236.

The sum of the mass numbers on the left should equal 236. 235 + 1 = 236

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3 years ago

A 35-year-old patent clerk needs glasses of 50-cm focal length to read patent applications that he holds 25 cm from his eyes. Fi

Natali5045456 [20]

Answer:

28.57 cm

Explanation:

We are given that

Focal length, $f=50 cm$

Distance of application from his eyes,s=40 cm

$\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}$

$\frac{1}{40}+\frac{1}{s'}=\frac{1}{50}$

$\frac{1}{s'}=\frac{1}{50}-\frac{1}{40}=\frac{4-5}{200}$

$s'=\frac{200}{-1}=-200 cm$

s=25 cm

Substitute the values

$\frac{1}{25}-\frac{1}{200}=\frac{1}{f'}$

$\frac{8-1}{200}=\frac{1}{f'}$

$\frac{1}{f'}=\frac{7}{200}$

$f'=\frac{200}{7}=28.57 cm$

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3 years ago