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2 years ago

A truck of mass 1800kg is moving with a speed 54km/h. When brakes are applied, it

Physics

1 answer:

EleoNora [17]2 years ago

6 0

Force = 3200 N

Work done = 640, 000 Nm

Explanation:

We begin by calculating the deceleration of the truck, using the velocity and distance;

a = (v² – u²)/2s

whereby;

a = acceleration

v = initial velocity

u = initial velocity

s = distance

We begin by changing the speed from km/h into m/s;

54km/hr = 15m/s

Then acceleration;

a = (0² – 15²) / 2 * 200

a = -225 / 400

a = - 0.5625 m/ s²

To calculate force;

F = ma

Whereby;

F = force

M = mass (in kgs)

a = acceleration

F = 1800 / 0.5625

F = 3200 N

Work done = Force * displacement

Work done = 3200 * 200

= 640, 000 Nm

Learn More:

For more on force and work done check out;

brainly.com/question/8662583

brainly.com/question/1268612

brainly.com/question/11870590

brainly.com/question/9125094

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1.024 × 10⁸ m

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2v₀ = Rω

substituting ω = v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

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So, R = 2 × 8RE

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2 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

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Explanation:

Given

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$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

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$\frac{M_x}{M_y}=0.05806\approx 0.0581$

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2 years ago

A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an

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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:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

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

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Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

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

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

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

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

=> $n = 31.6 \ moles$

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

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

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2 years ago

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Explanation:

Given

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Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

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2 years ago