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
madam [21]
10 months ago
9

how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?

Physics
1 answer:
Mars2501 [29]10 months ago
8 0

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

Explanation:

Typically, the mathematical formula for kinetic energy is as follows:

KE=\frac{1}{2} *m*v^{2}

K.E= kinetic energy

M=mass

V= speed

so,

KE=1/2*0.008*(420)*420

K.E=672.4J

 Therefore,

v*v=672.4/85

v*v=15.821

v=4.0m/s

Therefore

v=4.0m/s

What is Kinetic Energy?

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

To know more about Kinetic Energy? Visit to: -

brainly.com/question/22174271

#SPJ4

You might be interested in
Ill give brainliest! 1. Which item below describes a quick change to
stiks02 [169]
The best answer to go with is b
7 0
2 years ago
Read 2 more answers
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
Mars has two moons. What does the orbital speed of these moons depend on?
Radda [10]

Answer:

their masses and their distances from Mars

Explanation:

Hope this helped

Also can u pls mark me brainliest

8 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will
nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

6 0
2 years ago
Other questions:
  • What two forces are there when you skydive
    6·2 answers
  • you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
    7·1 answer
  • If the force applied to an object is
    15·1 answer
  • A car traveling at 10 m/s speeds up to 20 m/s in 3 seconds. determine the acceleration of the car
    8·2 answers
  • I WILL MARK BRAINLIEST IF CORRECT!!!
    5·1 answer
  • Explain resolution of Force​
    6·1 answer
  • An objest with the mass of 2 kg is accelerated at 4 m/s2. the net force acting on the object is
    10·1 answer
  • How are molecule shapes determined by electrons
    12·2 answers
  • Predict the magnitude of the gravitational force....
    6·1 answer
  • What is the best prediction for what will happen when all of the hydrogen in the sun has undergone fusion? helium nuclei will sp
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!