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2 years ago

11

A skier of mass 72 kg is pulled up a slope by a motor-driven cable. a. how much work is required to pull him 75 m up a 30 degree

slope (assumed frictionless) at a constant speed of 3.4 m/s

1 answer:

likoan [24]2 years ago

8 0

Answer: $26.460\times 10^3\ J$

Explanation:

Given the mass of skier m=72 kg

distance traveled d=75 m

constant speed v=3.4 m/s

If speed is constant then there must no force acting in the direction of motion

i.e. tension force must be equal to the component of weight

$T=mg\sin 30^{\circ}$

Work done is given by

$\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J$

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The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em

Vilka [71]

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

mass of filled can, $m_f=0.369\ kg$

weight of empty can, $w_c=0.153\ N$

<u>So, mass of the empty can:</u>

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

<u>Hence the mass of liquid(soda):</u>

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

<u>Therefore the density of liquid soda:</u>

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

<u>Specific weight of the liquid soda:</u>

$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$

3 0

2 years ago

Read 2 more answers

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

2 years ago

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

$V_t=\frac{4}{3} \pi.r_2^3$

And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

Therefore the net volume of material required to make the sphere:

$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

<u>Then the mass of the material used is:</u>

$m=\rho.V$

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

4 0

3 years ago

A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio

Maurinko [17]

The answer is D hoped this helped

4 0

2 years ago

A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the

levacccp [35]

Tension in the rope due to applied force will be given as

$F = 142 N$

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

$W = F d cos\theta$

now we can plug in all values above

$W = 142 * 6.1 * cos37$

$W = 691.8 J$

So here work done to pull is given by 691.8 J

8 0

2 years ago