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ivanzaharov [21]
3 years ago
15

A box is being pushed but two stellar science students, one on each side of the box. Peter is pushing box with a force of 10 N t

o the left. Josh is pushing the box with a force of 15 N to the right. Who is the stronger individual and what is the net force and direction on the box?
Physics
1 answer:
Vitek1552 [10]3 years ago
5 0

Answer:

Josh is the strongest

The net force is 5N towards right

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Cause and Effect: What would happen to a space vehicle in orbit around Earth if it sped up?
gulaghasi [49]

Answer:

G M m / R^2 = m v^2 / R     gravitational force = centripetal force

G M = v^2 R   =    constant

As v increases R will must decrease

Take the moon as an example

S = 2 pi R   where R is about 240,000 miles for one orbit

S / 1 day = 54,000 miles/day  for a 28 day circuit

S / 1 hr = 54000 / 24 = 2200 mph  which is much less than a satellite in orbit

4 0
2 years ago
If a 8.1g ring is heated using 10.0 ), its temperature rises 21.7°C. Calculate the specific
lakkis [162]
the water specific heat will remain at 4.184.
7 0
3 years ago
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

7 0
3 years ago
A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. What is the magnitude and direction of the
Nuetrik [128]

Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

<h3>What is displacement of a body in motion?</h3>
  • The displacement is the shortest distance between initial and final positions of a body.
  • It's a vector quantity, and can positive, negative, or zero.
  • The magnitude of displacement is less than or equal to the distance travelled.
<h3>How to solve the problem?</h3>
  • At first, we can draw a diagram showing the motion of the body.
  • From the diagram, the displacement of the body will be equal to the distance between point A and D.
  • To solve this, we can use Pythagoras theorem.

AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

Learn more about the Displacement here:

brainly.com/question/28020108

#SPJ4

3 0
2 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
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