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
bezimeni [28]
3 years ago
10

Why is it important to have a strong core (mid section)?

Physics
1 answer:
Reika [66]3 years ago
6 0

Answer:

Your core stabilizes your body, allowing you to move in any direction as well as having proper balance. It helps prevents falls and supports your body. So having a strong core is beneficial to everyone because it allows your body to function properly. Improved Balance and Stability.

You might be interested in
In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar
katrin2010 [14]

Time period of any moon of Jupiter is given by

T = 2\pi \sqrt{\frac{r^3}{GM}}

from above formula we can say that mass of Jupiter is given by

M = \frac{4 \pi^2 r^3}{GT^2}

now for part a)

r = 4.22 * 10^8 m

T = 1.77 day = 152928 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}

M = 1.9* 10^{27} kg

Part B)

r = 6.71 * 10^8 m

T = 3.55 day = 306720 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}

M = 1.9* 10^{27} kg

Part c)

r = 10.7 * 10^8 m

T = 7.16 day = 618624 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}

M = 1.89* 10^{27} kg

PART D)

r = 18.8 * 10^8 m

T = 16.7 day = 1442880 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}

M = 1.889* 10^{27} kg

6 0
3 years ago
The process of changing the energy of a system by means of force. Force * Diatance
Dmitry_Shevchenko [17]
This is known as work
7 0
3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of
kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0
3 years ago
How long does it take a car to cross a 20m bridge if it starts from rest and accelerates at 5 m/s^2?
polet [3.4K]

The correct answer is 2.8s

5 0
3 years ago
Other questions:
  • Which statement accurately describes the chemical symbol of an element?
    15·2 answers
  • A 500 μF capacitor is wired in series with a 5 V battery and a 20 kΩ resistor. What is the voltage across the capacitor after 20
    15·2 answers
  • A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
    12·1 answer
  • How much work does the electric field do in moving a proton from a point with a potential of +130 V
    9·1 answer
  • A bodybuilder loads a bar with 550 Newton’s (125 pounds) of weight and pushes the bar over her head 10 times. Each time she lift
    12·1 answer
  • What is the wavelength of an electromagnetic wave that travels at 3 10 m/s and has a frequency of 60 mhz? (1 mhz = 1,000,000 hz)
    7·2 answers
  • Iron is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron forms a cr
    9·1 answer
  • Which of the following is NOT a function of MITOSIS?
    15·1 answer
  • If a car travels 200 m to the east in 8.0s what is the car's average velocity.
    12·1 answer
  • Which substance may lower air temperatures after a volcanic eruption?
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!