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3 years ago

13

The law of conservation of energy and describe the energy transformation that occur as you coast down a long hill on a bicycle a

nd then apply the brakes to make the bike stop bottom

1 answer:

Phantasy [73]3 years ago

8 0

As you coast down a long hill on your bicycle, potential energy from your height is converted to kinetic energy as you and your bike are pulled downward by gravity along the slope of the hill. While there is air resistance and friction slowing you down by a little bit, your speed increases gradually until you apply the brakes, causing enough friction to slow yourself and the bike to a stop at the bottom

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During physical activity of high intensity, your heart will beat at a percentage of its maximum rate. Whi

Maurinko [17]

Answer:

D

Explanation:

If it is of very high intensity it will be 85-100%

8 0

3 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

4 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

3 years ago

What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N

VMariaS [17]

The gravitational force between Mars and the Sun is $1.65\cdot 10^{21} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the Sun

$m_2 = 6.39\cdot 10^{23} kg$ is the mass of Mars

$r=229\cdot 10^6 km = 229\cdot 10^9 m$ is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N$

So, the closest answer is

$1.65\cdot 10^{21} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

4 years ago

Which vector shows the direction of the centripetal acceleration at this point

Mademuasel [1]

Answer:

It's B

Explanation:

6 0

3 years ago