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3 years ago

A force of 900 N pushes a wedge 0.10 m into a log. If the work done on the log is 50 J, what is the efficiency of the wedge?

Physics

1 answer:

Anna [14]3 years ago

8 0

Answer:

55.56%

Explanation:

We'll begin by calculating the work done in pushing the wedge. This can be obtained as follow:

Force (F) = 900 N

Distance (s) = 0.1 m

Workdone (Wd) =..?

Wd = F × s

Wd = 900 × 0.1

Wd = 90 J

Finally, we shall determine the efficiency. This can be obtained as follow:

Work output = 50 J

Work input = 90 J

Efficiency =?

Efficiency = work output / work input

Efficiency = 50 / 90 × 100

Efficiency = 55.56%

Thus, the efficiency is 55.56%

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A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

6 0

3 years ago

A gas bottle contains 0.250 mol of gas at 730 mm hg pressure. if the final pressure is 1.15 atm, how many moles of gas were adde

Ludmilka [50]

Answer: 0.049 mol

Explanation:

1) Data:

n₁ = 0.250 mol

p₁ = 730 mmHg

p₂ = 1.15 atm

n₂ - n₁ = ?

2) Assumptions:

i) ideal gas equation: pV = nRT

ii) V and T constants.

3) Solution:

i) Since the temperature and the volume must be assumed constant, you can simplify the ideal gas equation into:

pV = nRT ⇒ p/n = RT/V ⇒ p/n = constant.

ii) Then p₁ / n₁ = p₂ / n₂

⇒ n₂ = p₂ n₁ / p₁

iii) n₂ = 1.15atm × 760 mmHg/atm × 0.250 mol / 730mmHg = 0.299 mol

iv) n₂ - n₁ = 0.299 mol - 0.250 mol = 0.049 mol

7 0

3 years ago

A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.

damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force = 20N

Compression = 0.08m

Unknown:

Spring constant = ?

Elastic potential energy = ?

Solution:

To solve this problem, we use the expression below:

F = k e

F is the force

k is the spring constant

e is the compression

20 = k x 0.08

k = 250N/m

Elastic potential energy;

EPE = $\frac{1}{2}$ k e² = $\frac{1}{2}$ x 250 x 0.08²

Elastic potential energy = 0.8J

3 0

3 years ago

a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0

3 years ago

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

3 years ago