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2 years ago

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A car is moving down a flat, horizontal highway at a constant speed of 21 m/s when suddenly a rock dropped from rest straight do

wn from a bridge crashes through the windshield. The rock strikes the windshield at a right angle, and the windshield makes a 60 degree angle with the horizontal.
a. Find the speed of the rock relative to the car when it hits the windshield.
b. Find the distance above the point of contact with the windshield from which the rock was dropped.
c. If the driver had seen what was about to happen just as the rock was being released, how much time would she have had to react before the rock strikes the windshield?

1 answer:

Naddik [55]2 years ago

7 0

Answer:

a)9.8

Explanation:

because any object falling from any height is 9.8

(sorry that's the only one i know)

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A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is th

Usimov [2.4K]

Answer:

$v_{f}$ = 3289.8 m / s

Explanation:

This exercise can be solved using the definition of momentum

I = ∫ F dt

Let's replace and calculate

I = ∫ (at - bt²) dt

We integrate

I = a t² / 2 - b t³ / 3

We evaluate between the lower limits I=0 for t = 0 s and higher I=I for t = 2.74 ms

I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)

I = a 3,754 - b 6,857

We substitute the values of a and b

I = 1500 3,754 - 20 6,857

I = 5,631 - 137.14

I = 5493.9 N s

Now let's use the relationship between momentum and momentum

I = Δp = m $v_{f}$ - m v₀o

I = m $v_{f}$ - 0

$v_{f}$ = I / m

$v_{f}$ = 5493.9 /1.67

$v_{f}$ = 3289.8 m / s

5 0

2 years ago

A person in a boat sees a fish in the water (n = 1.33, the light rays making an angle of 40? relative to the water's surface. wh

jek_recluse [69]

We know that the measure of an incident ray is: α 1 = 40°.
The index of refraction:
- for the air : n 1 = 1.00,
- for the water: n 2 = 1.33
Snell`s Law of Refraction :
n 1 · sin α 1 = n 2 · sin α 2
sin α 2 = n 1 · sin α 1 / n 2 =
= 1.00 · sin 40° / 1.33 = 0.64278 / 1.33 = 0.4833
α 2 = sin ^(-1) 0.4833
α 2 = 28.9 °
Answer: The angle relative to the water`s surface of the rays when beneath the surface is 28.9°.

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2 years ago

The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star

Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

$PV=nRT$

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3$

V=8.86*10^{50}L

and for n

$n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol$

Hence, we have (1 Pa = 9.85*10^{-9}atm)

$T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K$

hope this helps!!

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2 years ago

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago

A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal

ASHA 777 [7]

Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{68*3.9^2}{6.5}$
$F_{centripetal\:force} = \frac{68*15.21}{6.5}$
$F_{centripetal\:force} = \frac{1034.28}{6.5}$
$\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}$
Answer:
<span>B.159 N</span>

3 0

2 years ago