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natta225 [31]
3 years ago
13

True or False:

Physics
2 answers:
viva [34]3 years ago
8 0

Answer:

True. The object's location is based off of the reference point, so by moving the reference point, you move the object's location.

Explanation:

pantera1 [17]3 years ago
3 0

Answer:

True

Explanation:

the reference point is where the object starts so changing where it starts changes the objects location.

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In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?
Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

  • negative charges
  • positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

#SPJ1

5 0
1 year ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
How far did an airplane go if it traveled at 700 mph for 3.5 hours?
Scrat [10]
2,450 miles. you have to do 700•3.5= 2,450
6 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
2 years ago
A rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. Calculate the normal force (in N) acting on an a
goldenfox [79]

According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is  -6048 N

<h3>What does Newton's third law say ?</h3>

The law state that in every action, there will be equal and opposite reaction.

Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.

Let us first calculate the force involved in the acceleration of the rocket by using the formula

F = ma

Where mass m = 87.4 kg, acceleration a = 69.2 m/s2

Substitute the two parameters into the formula

F = 87.4 x 69.2

F = 6048.08 N

According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.

Therefore, the normal force acting on the astronaut is -6048 N approximately

Learn more about forces here: brainly.com/question/12970081

#SPJ1

7 0
2 years ago
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