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Elis [28]
3 years ago
14

Donny, who has a mass of 75.0 kg, is riding at 25.0 m/s in his truck when he suddenly slams on the brakes to avoid hitting a squ

irrel crossing the road. His seatbelt brings his body to a stop in 0.500 seconds. What force does the seatbelt exert on him?
Physics
1 answer:
marta [7]3 years ago
5 0

Answer:

F = 3750  N

Explanation:

Given that,

Mass of Donny, m = 75 kg

Initial speed, u = 25 m/s

He suddenly slams on the brakes to avoid hitting a squirrel crossing the road, final speed = 0

Time, t = 0.5 s

We need to find the force the seatbelt exert on him. The force is given by :

F = ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{75\times (0-25)}{0.5}\\\\=-3750\ N

So, the force is 3750  N.

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A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165
Anna007 [38]

Answer:

41.3 m/s^2 option (e)

Explanation:

force, F = 6.81 N

mass, m = 165 g = 0.165 kg

Let a be the acceleration of the puck.

Use newtons' second law

Force = mass x acceleration

6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

5 0
4 years ago
Please help will mark brainleist <br> ASAP
kiruha [24]

Answer:

B - Earth's path around the Sun

Explanation:

4 0
3 years ago
Read 2 more answers
g Power is _________________the force required to push something the work done by a system the speed of an object the rate that
Levart [38]

Answer:

the rate that the energy of a system is transformed

Explanation:

We can define energy as the capacity or ability to do work. Power is defined as the rate of doing work or the rate at which energy is transformed. It can also be regarded as the time rate of energy transfer. In older physics literature, power is sometimes referred to as activity.

Power is given by energy/time. Its unit is watt which is defined as joule per second. Another popular unit of power is horsepower. 1 horsepower = 746 watts.

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4 0
3 years ago
The measurement 0.025563 g should be reported as?
Tomtit [17]
The given mass is 0.025563 g.

Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.

b. 2.5 x 10² g = 250 g.
It is incorrect.

c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.

d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.

Answer: a.  0.026 g
7 0
3 years ago
Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p
Bad White [126]

Answer:

Diode Lasers  

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250  

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is  

4. The optical gain bandwidth (as measured between half intensity points) will  

normally depend on the pumping current (diode current) but for this problem  

assume that it is 2 nm.  

(a) What is the mode integer m of the peak radiation?  

(b) What is the separation between the modes of the cavity? Please express your  

answer as Δλ.  

(c) How many modes are within the gain band of the laser?  

(d) What is the reflection coefficient and reflectance at the ends of the optical  

cavity (faces of the InGaAsP crystal)?  

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction  

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the  

beam is a Gaussian beam with the waist located at the output. And the beam  

waist size is approximately the x-y dimensions of the cavity.)  

Solution:  

(a) The wavelength λ of a cavity mode and length L are related by  

n

mL

2

λ = , where m is the mode number, and n is the refractive index.  

So the mode integer of the peak radiation is  

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .  

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .  

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .  

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the  

bandwidth could fit in two possible modes.  

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0
4 years ago
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