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
Alecsey [184]
3 years ago
13

Muscles make up ore than __ of a person body?a.10b.25c40d.75

Physics
2 answers:
exis [7]3 years ago
8 0
40 percent of your body
Vsevolod [243]3 years ago
8 0
Yea it more then 40% C
You might be interested in
A rock falls off a cliff. How fast will it be going after falling for 4.33 seconds?
bixtya [17]

Answer:42.43m/s

Explanation:According to vf=vi+at, we  can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s

4 0
1 year ago
A plane lands on a runway with a speed of 115 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2
marin [14]

Answer:

<em> The planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

Acceleration can be represented mathematically as

a = v/t.................................... Equation 1

Where a = acceleration, v = velocity, t= time.

<em>Given: v = 115 m/s, t = 13.0 s</em>

<em>Substituting these values into equation 1</em>

<em>a = 115/13</em>

<em>a = 8.846 m/s² moving east</em>

<em>Thus the planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

4 0
3 years ago
Imagine a 10kg block moving with a speed of 20m/s. calculate the kinetic energy of this block
avanturin [10]
E_{c}=\frac{mv^2}{2}=2000J
7 0
3 years ago
Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as
djverab [1.8K]

Answer:

E=7.28\times 10^6\ J

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J

So, the gravitational potential energy of the person is 7.28\times 10^6\ J

6 0
3 years ago
A 70.0 kg skydiver falls towards the earth. If the force due to air resistance is 0 N, what is the acceleration of the skydiver?
lukranit [14]
The gravitational acceleration g = 9.81 m/s²
7 0
3 years ago
Read 2 more answers
Other questions:
  • What is "In an isolated system two cars, each with a mass of 2,000 kg, collide. Car 1 is initially at rest , while car 2 was mov
    9·1 answer
  • If a bat with a mass of 5 kg and acceleration of 2 m/s2 hits a ball whose mass is 0.5 kg in the forward direction, what is the r
    5·2 answers
  • 50 J of work was performed in 20 seconds. How much power was used to do this task?
    9·2 answers
  • A parallel-plate capacitor with plates of area 540 cm2 is charged to a potential difference V and is then disconnected from the
    7·1 answer
  • If a person is throwing darts at a target in shoots three in a row in the same spot on the 20 section outer ring, but none in th
    5·2 answers
  • Two resistors, R1=2.79 Ω and R2=6.37 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi
    7·1 answer
  • Two helium-filled balloons are released simultaneously at points A and B on the x axis in an earth-based reference frame. Balloo
    10·1 answer
  • An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel 1530 m/s in seawater. How deep is th
    15·1 answer
  • Plz answer fast the question
    5·1 answer
  • In terms of π, what is the length of an arc
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!