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3 years ago

An acrobat on skis starts from rest 50.0 m above the

Physics

1 answer:

Brums [2.3K]3 years ago

4 0

The initial speed of the skier is 14 m/s.

The maximum height attained by the skier is 15 m.

<h3>Initial speed of the skier</h3>

The initial speed of the skier is determined by using the following kinematic equation.

$K.E = mgh\\\\\frac{1}{2} mv^2 = mgh\\\\\frac{1}{2} v^2 = gh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 10} \\\\v = 14 \ m/s$

<h3>Maximum height attained</h3>

The maximum height attained by the skier is calculated as follows;

$H = \frac{v^2 sin^2\theta}{2g} \\\\H = \frac{14^2 \times (sin45)^2}{2 \times 9.8} \\\\H = 5 \ m$

Thus, the maximum height attained by the skier is (10 + 5) = 15 m.

Learn more about maximum height of projectile here: brainly.com/question/12870645

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When two objects of different masses, different temperatures, and different sizes are placed in thermal contact, energy will alw

denpristay [2]

Answer:

a) from the hotter object to the cooler object

Explanation:

temperature moves by conduction, which is associated with the movement of atoms or molecules and the always move from hight temperatures to lower temperatures to attain thermal equilinrium of the system.

so when two objects are placed together and have different temperatures then the system is not in thermal equilibrium and to attain it, temperature can only move to coller object and not from the coller object according to thermodynamics.

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3 years ago

A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal

Margarita [4]

The electrostatic force between two charges is given by Coulomb's law:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
$F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N$

7 0

4 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Why is temperature scalar?

shepuryov [24]

Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.

But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).

3 0

3 years ago

Which statement best describes perigee?

pantera1 [17]

Answer:

A. The closest point in the Moon's orbit to Earth

Explanation:

The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.

Let's see instead the names of the other options:

B. The farthest point in the Moon's orbit to Earth --> this point is called apogee

C. The closest point in Earth's orbit of the Sun --> this point is called perihelion

D. The Sun's orbit that is closest to the Moon --> this point has no specific name

8 0

3 years ago

Read 2 more answers