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Tresset [83]
3 years ago
12

Coming to the bottom of a mountain, a skier moving with speed v collides with a barrier and is brought to a stop in an amount of

time t. Over the course of the impact, the skier experienced an average net force F from the barrier. If the skier\'s initial velocity had been the same (v), but the time of impact to come to a full stop had been four times as long (4t), what would have been the average net force exerted on the skier during the collision? The average net force would have been...a) F/4b)Fc)4Fd) F/2e)2FIf the time of impact to come to a full stop had been the same (t), but the skier\'s initial velocity had been half as much (v / 2), what would have been the average net force exerted on the skier during the collision? The average net force would have been...a) 4Fb)Fc)2Fd) F/4e)F/2
Physics
1 answer:
IRISSAK [1]3 years ago
8 0

Answer:

Explanation:

Impulse of a force is measured by force x time or F X t

Impulse also equals change in momentum or

F x t = m v₂ - m v₁

The given case is as follows

in the first case

F x t = mv - o = mv

F = mv / t

in the second case

F₁ x 4 t = mv

F₁ = 1/4 x mv /t

F₁ = F / 4

option a) is correct .

iii )

In the last case

F₂ X t = m v/2 -0

F₂ = 1/2 x mv / t

= 1/2 x F

F₂ = F/2

Option e ) is correct.

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A star's transverse velocity depends on which two factors?
worty [1.4K]
Distance and proper motion
8 0
3 years ago
A car starts from 0 m along a road and accelerates at 0.5 m/s^2 to the right. A second car starts from 1000 m along the road and
Brrunno [24]

Answer:

e) 31.6 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s_1=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_1=\frac{1}{2}0.5t^2\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow s_2=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_2=\frac{1}{2}1.5t^2\ m

s_1+s_2=1000

\\\Rightarrow 1000=\frac{1}{2}0.5t^2+\frac{1}{2}1.5t^2\\\Rightarrow 1000=\frac{0.5t^2+1.5t^2}{2}\\\Rightarrow 1000=\frac{2t^2}{2}\\\Rightarrow 1000=t^2\\\Rightarrow t=\sqrt{1000}\\\Rightarrow t=31.6\ s

Time taken by the cars to meet 31.6 seconds.

5 0
4 years ago
Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks
babymother [125]

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

H=Hi+Viy*t+1/2*g*t

X=Xi+Vx*t

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

t=(5.40m)/(3.50m/s)

Then t=1,54s

After that we need to replace t to find H, this way

H=(1/2)*(9,81m/s^2)*(1,54s)^2

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

t=(5,40m)/(2*3,50m/s)

Then, t=0,77

4 0
3 years ago
For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m
Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

\Sigma F = F-R = 0 (Eq. 1)

Where:

F - Forward trust, measured in newtons.

R - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: (F = 6.85\times 10^{5}\,N)

R = F

R = 6.85\times 10^{5}\,N

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

4 0
4 years ago
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after
wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0
3 years ago
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