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3 years ago

13

A convex lens bulges outward in its center. and it refracts light rays towards its center. Which represents a convex lens? A) A

B) B C) A and B D) neither A nor B

2 answers:

vlada-n [284]3 years ago

7 0

Answer:

A

Explanation:

Mila [183]3 years ago

4 0

Answer:

Its A

Explanation:

The lens bulges outward in its center, like reading glasses.

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A child wearing socks is trying to see how far he can slide across his kitchen floor. If

Ugo [173]

Answer:

The coefficient of static friction between the ground and the soles of a runner’s shoes is 0.98. What is the maximum speed in which the runner can accelerate without slipping if they have a mass of 73 kg?

Explanation:

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2 years ago

An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu

Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

= e × V

= (1.6 × 10⁻¹⁹ C)×(1.2 V)

= 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

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2 years ago

Which is the mode of the following data?<br>120, 125, 130, 125, 135

Aleks [24]

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

8 0

3 years ago

Read 2 more answers

Heated air moves from baseboard heaters to the rest of a room in a process called

pychu [463]

It is called convection

3 0

3 years ago

PHYSICS, HELP PLZ??!! 100PTS

romanna [79]

Hi there!

We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:

$t_{max} = \frac{vsin\theta}{g}$

Where:

tmax = (? sec)

vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)

g = acceleration due to gravity (9.8 m/s²)

We can solve for this time:

$t_{max} = \frac{7.43}{9.8} = 0.758 s$

When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.

We must begin by solving for the maximum height reached by the ball using the equation:

$d = y_0 + v_{0y}t + \frac{1}{2}at^2$

d = displacement (m)

vi = initial velocity (7.43 m/s)

a = acceleration due to gravity

d = displacement (m)

y0 = initial VERTICAL displacement (28m)

Plug in the values:

$d = 28 + 7.43(0.758) + \frac{1}{2}(-9.8)(0.758^2) = 30.817 m$

Now, we can use the rearranged kinematic equation:

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(30.817)}{9.8}} = 2.51 s$

Add the two times together:

$0.758 + 2.51 = \boxed{3.266 s}$

7 0

2 years ago