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Dmitry_Shevchenko [17]
3 years ago
8

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s

in 0.820 s. (a) What is the magnitude of the linear impulse experienced by a 66.0-kg passenger in the car during the time the car accelerates?
Physics
1 answer:
yarga [219]3 years ago
6 0

Answer:

I=336.6kgm/s

Explanation:

The equation for the linear impulse is as follows:

I=F\Delta t

where I is impulse, F is the force, and \Delta t is the change in time.

The force, according to Newton's second law:

F=ma

and since a=\frac{v_{f}-v_{i}}{\Delta t}

the force will be:

F=m(\frac{v_{f}-v_{i}}{\Delta t})

replacing in the equation for impulse:

I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)

we see that \Delta t is canceled, so

I=m(v_{f}-v_{i})

And according to the problem v_{i}=0m/s, v_{f}=5.10m/s and the mass of the passenger is m=66kg. Thus:

I=(66kg)(5.10m/s-0m/s)

I=(66kg)(5.10m/s)

I=336.6kgm/s

the magnitude of the linear impulse experienced the passenger is 336.6kgm/s

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Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

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A helium-neon laser emits light of wavelength 632.8 nm and has a power output of 16 mw. how many photons are emitted per second
NemiM [27]
The power of the laser is
P=16 mW=0.016 W
by using the relationship between power, energy and time, we can find the energy delivered by the laser in 1 second:
E=Pt=(0.016 W)(1 s)=0.016 J

The frequency of the photons of this light is given by
f= \frac{c}{\lambda}= \frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m} =4.74 \cdot 10^{14} Hz
and the energy of a single photon is
E_1=hf=(6.6\cdot 10^{-34} Js)(4.74 \cdot 10^{14} Hz)=3.13 \cdot 10^{-19} J

In order to find the number of photons emitted per second, we must divide the total energy emitted by the laser by the energy of a single photon, and we get:
N= \frac{E}{E_1}= \frac{0.016 J}{3.13 \cdot 10^{-19}J} =5.11 \cdot 10^{16} photons
7 0
3 years ago
One of the greatest terrorism-related nuclear threats is from Select one: a. nuclear power plants. b. dirty bombs. c. nuclear wa
liq [111]

Answer:

Dirty bomb

Explanation:

Among the nuclear bomb One type is a "dirty bomb." It combines a conventional explosive such as the dynamite with radioactive material which can spread when the system explodes.  The explosion is releasing "dirty" bits of radioactive particles which are extremely harmful and can cause loss equivalent to a nuclear attack.

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7 0
2 years ago
Wooden ball lies in a vessel with water so that half of it is immersed in water.
Ugo [173]

Downward force acting on the ball is 19.6N

Net force acting on the ball is 1960V N

<u>Explanation:</u>

<u />

Given:

Mass of the ball, m = 2kg

Density of ball, σ = 800 kg/m³

Density of water, ρ = 1000 kg/m³

Downward force acting by the ball in the vessel = mg

where, g = 9.8m/s²

F = 2 X 9.8

F = 19.6N

Net force acting on the ball:

Fnet = (ρ - σ) Vg

where,

V is the volume of water

Fnet = (1000 - 800) V X 9.8

Fnet = 1960V N

If the volume is known, then substitute the value of V to find the net force.

Thus, Downward force acting on the ball is 19.6N

         Net force acting on the ball is 1960V N

5 0
4 years ago
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