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2 years ago

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in:

1 answer:

4 0

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in: 2 seconds

Answer:

2.4 m

Explanation:

From the question above,

Applying equation of motion,

s = ut+at²/2....................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance.

make s the subject of the equation,

Given: a = 1.8 m/s², t = 2 seconds, u = 0 m/s (from rest)

Substitute these value into equation 1

s = 0(2)+1.8(2²)/2

s = 1.2(4)/2

s = 2.4 m.

Hence she has traveled 2.4 m

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2 years ago

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A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f

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Answer:

$2.5 \times 10^{11} N-m$ upwards

Explanation:

Torque is the vector cross product of the force and radial distance.

$\tau = rF sin \theta$

$\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m$

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2 years ago

Calculate the pressure, in atmospheres, exerted by each of the following:

gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

P = 2.15 × 0.08206 × 330/55

P = 1.06 atm

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2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

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2 years ago

An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac

Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

= (50 - 0)/4

= 50/4 = 12.5 (m/s2)

Hope this helps!

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2 years ago