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2 years ago

A 70.0 kg skier is at rest at the top of a 120m hill. Assuming friction is negligible.

Physics

1 answer:

jasenka [17]2 years ago

6 0

Answer:

49 m/s

Explanation:

Initial potential energy = final kinetic energy

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 10 m/s² × 120 m)

v = 49 m/s

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A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

$v = 70 m/s$

now we have

$v_x = 70 cos25$

$v_x = 63.44 m/s$

$v_y = 70 sin25$

$v_y = 29.6 m/s$

now in Y direction we can use kinematics

$v_y = v_i + at$

$v_y = 29.6 - (9.81 \times 5)$

$v_y = -19.5 m/s$

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{63.44^2 + 19.5^2}$

$v = 66.4 m/s$

4 0

2 years ago

a glass vessel is completely filled with 340 gram of water at zero degree celsius what weight of Mercury will overflow when the

Svetlanka [38]

Answer:

A glass flask whose volume is 1000 cm ^3 at 0.0 ^oC is completely filled with mercury at this. Every substance when heat energy is supplied, expands due to the Rate of thermal expansion will be different for different materials. Volume of the glass flask and mercury at 0 degree Celsius V0=1000cm3=1×10−3m3 V 0

Explanation:

hope dis help!!!

6 0

2 years ago

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

2 years ago

PLS ANSWER ASAP!!

Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the poles

Mathematically:

FM = K $\frac{M1 M2}{r^{2} }$

here r is the distance between the poles

FM is the magnetic force between the poles

M1 is the strength of the first magnetic pole

M2 is the strength of the second pole

K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0

2 years ago

Soot particles ("black carbon aerosols" generally cause ________ of earth's atmosphere by ________ solar energy.

sineoko [7]

They cause an increase in temp of earths atmosphere or warming by absorbing solar energy. hope this helps

7 0

2 years ago