How many significant figures are in 73,102L

If a hammer has a resistance arm of 3 inches and the effort arm is 15 inches. What is the mechanical advantage?

Sunny_sXe [5.5K]

Answer:

Option D

MA=5

Explanation:

Mechanical advantage is the the ratio of the output force to the input force hence

$MA=\frac {Effort arm length}{Load arm length}$

Substituting 15 inches for the effort arm length and 3 inches for the load arm length then we obtain the mechanical advantage as

$MA=\frac {15}{3}=5$

4 0

3 years ago

Bill is throwing a football at four targets and attempting to knock them over. Which of the following targets will be hardest fo

Naddika [18.5K]

100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.

7 0

3 years ago

You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
<span> $h=vt_2=(343 m/s)(1.13 s)=388 m$ </span>

3 0

3 years ago

Read 2 more answers

A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0

2 years ago

If two swimmers compete in a race, does the faster swimmer develop more power?

valkas [14]

Power is equal to energy per unit time. In this case, power is proportional to energy while is inversely proportional to time,on the other hand. Given the two swimmers exerts same amount of energy but the faster swimmer just does things in faster time, then the faster swimmer should develop more power from shorter time

7 0

3 years ago