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larisa86 [58]
2 years ago
14

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this

first snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 16.1 m/s. The first one is thrown at an angle of 72◦ with respect to the horizontal. At what angle should the second snowball be thrown to arrive at the same point as the first? Answer in units of ◦
Physics
1 answer:
allsm [11]2 years ago
8 0

Answer:

\theta_2 \approx 40.5^{\circ}

Explanation:

<u>Given:</u>

velocity of the ball first and ball second, v_1=v_2=v\,m.s^{-1}

angle of projection of the first ball, \theta_1= 72^{\circ}

∵The balls should land at the same point,

∴their range of projectile,

R_1=R_2=R\,m

As we know for the range of projectile:

R=\frac{v^2}{g}.sin\,2\theta

∵ we have equal range in both the cases

\therefore \frac{v_1^2}{g}.sin\,2\theta_1=\frac{v_2^2}{g}.sin\,2\theta_2

\Rightarrow \frac{v^2}{g}.sin\,(2\times 72) =\frac{v^2}{g}.sin\,2\theta_2

\Rightarrow sin 144^{\circ}=sin\,2\theta_2

2\theta_2=sin^{-1}[sin 144^{\circ}]

\theta_2=\frac{1}{2}\times sin^{-1}[sin 144^{\circ}]

\theta_2 \approx 40.5^{\circ}

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Metal ores

Explanation:

in an area where subduction has occurred in times past, metal ores are likely to be found.

Metallic ores find subduction zone regions very favorable to crystallize out of a magma.

  • Ores have different modes of formation.
  • Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
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At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.

Learn more:

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4 0
3 years ago
Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:
White raven [17]

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

7 0
3 years ago
The way to earn a college diploma is by?
Morgarella [4.7K]
Going to college and passing all your classes
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A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
What are the two factors that affect the friction force between two surfaces
Masja [62]

Answer:

coefficient of static friction of the surface and the normal force

Explanation:

The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR

5 0
3 years ago
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