1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dimulka [17.4K]
3 years ago
6

I NEED HELP, PLEASE.

Physics
1 answer:
Zinaida [17]3 years ago
5 0
Hello,


I'd be glad to help.

Momentum  =  (mass) times (speed)

                                           =  (75 kg)  x  (18 m/s)

                                           =         1,350 kg-m/s  .    

Hope this helps
You might be interested in
Help with these three
Elenna [48]
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
g = 9.81 \frac{m}{s {}^{2} }
let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} }  \\ f = 325n
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
8 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
1.)State which definition of power below is correct.
Natalija [7]

C

kW, watts, kilowatts, W

6 0
3 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
Enumerate the viscosity of magma in different conditions.
Firlakuza [10]

Answer:

The nature of volcanic eruptions is highly dependent on magma viscosity and also on dissolved gas content. ... long it takes the treacle to flow from one end of a boiling tube to the other.

3 0
3 years ago
Other questions:
  • Keaton is asked to solve the following physics problem:
    6·1 answer
  • A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co
    9·1 answer
  • Penny has two dogs: Fluffy (left) and Butch (right).
    11·2 answers
  • Which statement describes the difference between speed and velocity
    8·2 answers
  • The law of _________ states that energy is not truly produced, but really only manipulated from one form to another.
    8·1 answer
  • The width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature acros
    10·1 answer
  • Define fundamental unit.​
    14·1 answer
  • A Projectile attains its maximum height when the angle of projection is?​
    7·1 answer
  • Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift dependi
    14·1 answer
  • A weather balloon has a volume of 35 L at sea level (1.0 atm). After the balloon is released it rises to where the air pressure
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!