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3 years ago

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A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on

the boy?

1 answer:

telo118 [61]3 years ago

8 0

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

$sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m$

The work done by gravity on the boy is calculated as;

W = P.E = mgh

= 32kg x 9.8m/s² x 1.928m

= 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

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A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of

mote1985 [20]

Answer:

<h2>640N/cm^2</h2>

Answer D is correct

Explanation:

$pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640$

hope this helps

brainliest appreciated

good luck! have a nice day!

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In which of the following situations is the Doppler effect absent?

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The source and the observer are moving towards each other. The observer is moving toward the source. The source is moving away from the observer

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Do humans have a gravitational field if so, how strong

lilavasa [31]

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Now, if you have mass, then you have weight ... it's the answer you give when somebody asks "How much do you weigh ?". That's the force of gravity between you and the Earth, pulling you toward the center of the Earth. But it doesn't stop there. There's also a force of gravity pulling the Earth toward the center of you. The strength of it is EQUAL to your weight.

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Read 2 more answers

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.

UNO [17]

Explanation:

The given data is as follows.

Electric field strength (E) = 187,500 V/m

Magnetic field strength (B) = 0.125 T

Diameter (d) = 25.05 cm = 0.2505 m (as 1 m = 100 cm)

Radius (r) = $\frac{d}{2}$

= $\frac{0.2505}{2}$

= 0.12525 m

Formula to calculate the magnetic force ( $F_{M}$ ) is as follows.

$F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

$F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

$F_{M} - F_{E}$ = 0

or, $F_{M} = F_{E}$

Hence, from equations (1) and (2)

Bqv = qE

or, v = $\frac{E}{B}$ ............. (3)

= $\frac{187500 V/m}{0.125 T}$

= 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

$F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where, $F_{c}$ = centripetal force

$F_{M} = F_{c}$

Using equation (1) and (4) as follows.

$F_{M} = F_{c}$

Bqv = $\frac{mv^{2}}{r}$

$\frac{q}{m} = \frac{v}{Br}$

= $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

= $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$ .

8 0

3 years ago

slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular

n200080 [17]

Answer:

a) $\%V = 87.36\,\%$ , b) $x = 1.248\,m$ , c) $F_{B} = 176488.341\,N$ , d) Six polar bears.

Explanation:

a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:

$\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0$

The height of the ice submerged is:

$\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V$

$x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}$

$x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}$

$x = 1.248\,m$

The percentage of the volume of the ice that is submerged is:

$\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%$

$\%V = 87.36\,\%$

b) The height of the portion of the ice that is submerged is:

$x = 1.248\,m$

c) The buoyant force acting on the ice is:

$F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F_{B} = 176488.341\,N$

d) The new system is modelled after the Archimedes' Principle and Newton's Laws:

$\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0$

The number of polar bear is cleared in the equation:

$n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V$

$n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}$

$n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}$

$n = 6.5$

The maximum number of polar bears that slab could support is 6.

8 0

3 years ago