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2 years ago

Select the correct answer

Physics

1 answer:

VMariaS [17]2 years ago

4 0

Answer:

Mary's average speed is 0.42 meters/second.

Explanation:

Mary's total travel was 250 meters and the elapsed time was 10 minutes, equivalent to 600 seconds.

d = displacement = 80 meters + 125 meters + 45 meters = 250 meters

t = time = 10 minutes . 60 seconds/minute = 600 seconds

We can obtain now the average speed:

s = average speed = displacement / time = 250 meters / 600 seconds ≅ 0.4167 ≅ 0.42

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In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/

polet [3.4K]

V o - initial velocity
v = velocity at the maximum height,
v² = v o² - 2 g h
v = 0
0 = v o² - 2 g h
v o² = 2 g h = 2 · 9.80 · 0.460
v o² = 9.052
v o = √9.052 = 3.004197 m/s ≈ 3 m/s

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3 years ago

An underwater scuba diver sees the Sun at an apparent angle of 43.0° above the horizontal. What is the actual elevation angle of

Inessa05 [86]

Answer:

The actual elevation angle is 12.87 degrees

Explanation:

In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that $\theta_2=90-43=47$ .

Using Snell's Law we can write:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2)$ ,

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

$\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)$

$\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13$ . Calling $\alpha$ the actual angle of elevation, we get from the picture that $\alpha=90-77.13=12.97$

7 0

3 years ago

Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you

garik1379 [7]

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

Explanation:

The gravitational attraction is given by Newton law of gravitation as follows;

$Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}$

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

M = The mass of Jupiter

m = The mass of the nearby body

R = The distance between the centers of Jupiter and the body

From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies

Therefore, as one moves away, R increases, and the strength of gravity reduces

Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.

7 0

2 years ago

A body starts from rest and travels ‘s’ m in 2nd second, than acceleration is.

Lana71 [14]

Answer:

Explanation:

3 0

2 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago