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2 years ago

TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!

Physics

2 answers:

JulsSmile [24]2 years ago

8 0

Answer:

Explanation:

<u>Constant Speed Motion </u>

An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

It's required to find the distance traveled by someone walking at v=1.7 m/s for t=15.9 s. Substituting in the last equation:

d = 1.7 m/s * 15.9 s

d = 27.03 m

Diano4ka-milaya [45]2 years ago

3 0

The answer is 27.03 I just multiplied the two numbers

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Which of the following is a force or motion that

faltersainse [42]

Answer: this answer is D. That's what I think it is

Explanation:

7 0

2 years ago

1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational

enyata [817]

Answer:

d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

x = vox t

Y Axis

y = $v_{oy}$ t - ½ g t²

It's displacement is

d = x i ^ + y j ^ + z k ^

Substituting

d = (23 i ^ + 33 j ^ + 0) m

Using your notation

d = <23, 33, 0> m

The force of gravity is the weight of the body

W = m g

W = 1 9.8 = 9.8 N

In vector notation, in general the upward direction is positive

W = (0 i ^ - 9.8 j ^ + 0K ^) N

W = <0, -9.8, 0>

Work is defined

W = F. dy

W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

Cos 180 = -1

W = -F y

W = - 9.8 (33-0)

W = -323.4 J

6 0

2 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

$\eta=0.52$

Efficiency = 52%

8 0

3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$