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anygoal [31]
1 year ago
8

A 1000.–kilogram car traveling 20.0 meters per second east experiences an impulse of 2000. newton • seconds west. What is the fi

nal velocity of the car after the impulse has been applied?
Physics
1 answer:
emmainna [20.7K]1 year ago
5 0
Impulse is equal to change in momentum. So if impulse is 2000 then to solve for new velocity we just set it equal to equation for momentum.

First find original momentum by p=mv
p=1000*20=20000

So then taking that value minus the impulse since it was in opposite direction of original momentum it will slow it down some. To find new velocity we just take

20000-2000=18000=mv

v=18000/1000 =18m/s

Hope this helps!! Any questions please ask!!
Thank you!
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Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 36.5 cm while travelin
tia_tia [17]

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula  v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

  • radio waves are electromagnetic waves and as such they have the speed of light which is 3*10^{8}m/s.
  • also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.
  • calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

f=8.219*10^{8}Hz

7 0
2 years ago
Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee
valina [46]

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

3 0
1 year ago
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

7 0
2 years ago
The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0
2 years ago
Quinn is testing the motion of two projectiles x and y by shooting them from a sling shot. What can we say best describes the mo
Studentka2010 [4]

Explanation:

            A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path  subjected to gravity force only.

           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

             In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

3 0
1 year ago
Read 2 more answers
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