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2 years ago

A student wishes to work out how much power she uses to lift her body when climbing a

Physics

1 answer:

Bess [88]2 years ago

4 0

Answer

i'm not 100% sure but 1764

Explanation:

Work done = gravitational potential energy

Gravitational potential energy = mass(kg) × height(m) × gravitational field strength(N/kg)

We can assume that the student is on earth so the gravitational field strength is 9.8N/kg

So work done = 60 × 3×9.8

=1764

(if you need help calculating power but if you do just divide your answer by 12 and you will get 147)

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Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh

GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=N_1-mg=N_1-690$

Now, from Newton's second law:

$F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)$

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=mg-N_2=690-N_2$

Now, from Newton's second law:

$F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)$

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

$Difference=N_1-N_2=862.5-517.5=345\ N$

Therefore, the difference between the up and down scale readings is 345 N.

4 0

3 years ago

A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing.

svetlana [45]

Answer:

v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

P₁-P₂ = ρ g (v₂² - v₁²)

pressure is defined by

P = F / A

we substitute

ΔF / A = ρ g (v₂² - v₁²)

v₂² = $\frac{\Delta F}{A \ \rho \ g} + v_1^2$

suppose that the area of the wing is A = 1 m²

we substitute

v₂² = $\frac{1000}{1 \ 1.29 \ 9.8} + 63^2$

v₂² = 79.10 + 3969

v₂ = √4048.1

v₂ = 63.62 m / s

7 0

3 years ago

A boat is rowed perpendicular to the shore of a river that flows at 3.0 m/s as shown in the

vazorg [7]

The magnitude is 5 ms^-1

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The _______ is the temperature to which air must be cooled at constant pressure to reach saturation.

Dafna1 [17]

The answer is the dew point

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The order of elements in the periodic table is based on the atomic number.<br> true or false

kirill [66]

True the elements are ordered in the atomic number

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3 years ago