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2 years ago

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A 72 kg skydiver is descending on a parachute. His speed is still increasing at 1.2 m/s2. What are the magnitude and direction o

f the force of the parachute harness on the diver? What are the magnitude and direction of the net force on the diver?

1 answer:

frozen [14]2 years ago

8 0

Answer:

86.4 N downward

Explanation:

Force: This can be defined as the product of mass and acceleration of a body.

The S.I unit of Force is Newton(N).

The Expression of force is given as,

F = ma ................ Equation 1

Where F = force of the parachute harness, m = mass of the skydiver, a = acceleration of the skydiver.

Given: m = 72 kg, a = 1.2 m/s²

Substitute into equation 1

F = 72(1.2)

F = 86.4 N down ward.

Hence the force on the parachute harness = 86.4 N downward

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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

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Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

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A projectile was launched horizontally with a velocity of 468 m/s, 1.86 m above the ground. Calculate how long it would take for

elena-14-01-66 [18.8K]

Answer:

0.6 seconds

Explanation:

The time to fall from height h is ...

t = √(2h/g)

t = √(2(1.86 m)/(9.8 m/s^2)) ≈ √0.3796 s ≈ 0.616 s

It would take about 0.6 seconds for the projectile to hit the ground.

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2 years ago

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If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in

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For the answer to the question above,
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
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2 years ago

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

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2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

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Explanation:

From the question we are told that

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The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

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So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

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2 years ago