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Lubov Fominskaja [6]

3 years ago

14

Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe

r.)
a. The force is exerted horizontally while pushing the crate 5.20 m.
b. The force is exerted at an angle of 41.0

1 answer:

telo118 [61]3 years ago

6 0

Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

Given

Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

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Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

3 years ago

Which of the following is not a natural resource? a. time b. water c. land d. air Please select the best answer from the choices

AleksAgata [21]

Answer: Option (a) is the correct answer.

Explanation:

Resources which are created by human beings are known as man-made resources.

For example, glass, rayon, nylon etc are all man-made resources.

Whereas resources which are naturally created are known as natural resources.

For example, wind, air, water etc are all natural resources.

Thus, we can conclude that out of the given options time is not a natural resource.

3 0

3 years ago

A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the o

IgorLugansk [536]

Answer:

Explanation:

Using conservation of momentum

$m_1=59+220=279kg$

$m_2=2.1kg$

$U_1=12m/s$

$U_2=-44m/s$

$m_1U_1+m_2U_2=(m_1+m_2)V\\\\\frac{278\times 12-2.1\times 44}{279+2.1}=V\\\\V=11.58m/s$

6 0

3 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

3 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago