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vladimir1956 [14]

3 years ago

14

How much work is done by a crane that lowers 1000 N of material a distance of 150 m?

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

Force= 1000 Newtons. Distance= 150 metres. Work= 150000 Joules. So, the work done by crane that lowers 1000 Newton's of a material a distance of 150 meters is 150000 Joules.

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Bone has a Young's modulus of about

saveliy_v [14]

Answer: 4.74 mm

Explanation:

We can solve this problem with the following equation:

$Y=\frac{stress}{strain}$ (1)

Where:

$Y=1.8(10)^{10} Pa$ is the Young modulus for femur

$stress=\frac{F}{A}=1.58(10)^{8} Pa$ is the stress (force $F$ applied per unit of transversal area $A$ ) on the femur

$strain=\frac{\Delta l}{l_{o}}$

Being:

$\Delta l$ the compression the femur can withstand before breaking

$l_{o}=0.54 m$ is the length of the femur without compression

Writing the data in equation (1):

$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l_{o}}}$ (2)

$1.8(10)^{10} Pa=\frac{1.58(10)^{8} Pa}{\frac{\Delta l}{0.54 m}}$ (3)

Isolating $\Delta l$ :

$\Delta l=\frac{(1.58(10)^{8} Pa)(0.54 m)}{1.8(10)^{10} Pa}$ (4)

$\Delta l=0.00474 m$ (5) This is the compression in meters

Converting this result to millimeters:

$\Delta l=0.00474 m \frac{1000 mm}{1 m}=4.74 mm$

4 0

4 years ago

What is the velocity of an electron that has a de Broglie wavelength approximately the length of a chemical bond? Assume this le

melomori [17]

Answer : The velocity of an electron is, $6.1\times 10^{6}m/s$

Explanation :

According to de-Broglie, the expression for wavelength is,

$\lambda=\frac{h}{p}$

and,

$p=mv$

where,

p = momentum, m = mass, v = velocity

So, the formula will be:

$\lambda=\frac{h}{mv}$ .............(1)

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $1.2\times 10^{-10}m$

m = mass of electron = $9.11\times 10^{-31}kg$

v = velocity of electron = ?

Now put all the given values in formula 1, we get:

$1.2\times 10^{-10}m=\frac{6.626\times 10^{-34}Js}{(9.11\times 10^{-31}kg)\times v}$

$v=6.1\times 10^{6}m/s$

Thus, the velocity of an electron is, $6.1\times 10^{6}m/s$

6 0

4 years ago

What are some types of landforms on Earth’s surface?<br><br><br><br> PLS ANSWER QUICK 11 POINTS

Brrunno [24]

Answer:

plateau, mountains, hills, plains

5 0

3 years ago

Read 2 more answers

Calculate the recoil velocity in the horizontal direction, in meters per second, of a 1.25-kg plunger that directly interacts wi

mart [117]

Answer:

$v_1=-8.19\ m/s$ '

Explanation:

It is given that,

Mass of the plunger, $m_1=1.25\ kg$

Mass of the bullet, $m_2=0.0175\ kg$

Initially both plunger and the bullet are at rest, $u_1=u_2=0$

Final speed of the bullet, $v_2=585\ m/s$

Let $v_1$ is the final speed of the plunger. Using the conservation of momentum to find it. The equation is as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Since, $u_1=u_2=0$

$m_1v_1+m_2v_2=0$

$v_1=-\dfrac{m_2v_2}{m_1}$

$v_1=-\dfrac{0.0175\times 585}{1.25}$

$v_1=-8.19\ m/s$

So, the recoil velocity of the plunger is 8.19 m/s. Hence, this is the required solution.

7 0

3 years ago

As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.11-m-

soldier1979 [14.2K]

Answer:

The strength of the magnetic field is 0.0842 mT

Explanation:

Given:

Velocity of rod $v = 3.07$ $\frac{m}{s}$

Length of rod $l = 1.11$ m

Induced emf across the rod $\epsilon = 0.287 \times 10^{-3}$ V

According to the faraday's law

We have a special case for moving rod in magnetic field.

Induced emf in moving rod is given by,

$\epsilon = Blv$

Where $B =$ strength of magnetic field

$B = \frac{\epsilon}{lv}$

$B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}$

$B = 0.0842 \times 10^{-3}$ T

$B = 0.0842$ mT

Therefore, the strength of the magnetic field is 0.0842 mT

7 0

3 years ago