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yanalaym [24]
2 years ago
7

At the end of the school day, at exactly 2:30 pm, a group of students run out of the school building and reach the edge of the s

chool property at 2:30:45 s. Which of the following correctly describes the motion in terms of time?
Question 2 options:

A)

t1= 0, t2 = 2:30:45 s, ∆t = 45 s

B)

t1= 2:30, t2 = 45 s

C)

∆t = 2:30

D)

t2= 2:30:45 s, ∆t = 45 s
Physics
1 answer:
Fofino [41]1 year ago
3 0

Explanation:

Initial time, t₁ = 2:30 pm

Final time, t₂ = 2:30:45

We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.

Change in time,

\Delta t=t_2-t_1\\\\\Delta t=2:30:45-2.30\\\\\Delta t=45\ s

Hence, this is the required solution.

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An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

\ (x_{+ve}) = \frac{1}{2} \times 15 \times  20 \\\\

          = \frac{1}{2} \times 300 \\\\=  150 \\\\

Calculating the value of negative displacement upon the time t:

(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\

          = \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\

\to X= X_{+ve} + X_{-ve} \\\\

\to  150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25

That's why its value lie in "between 20 s and 30 s".

6 0
2 years ago
The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the
otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

8 0
2 years ago
Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No
Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field  

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns  = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0
2 years ago
Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller
victus00 [196]

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

mgH and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

a = \frac{v_f - v_i}{\Delta t}

since here initial and final speeds are same so they both must have same average acceleration here.

5 0
2 years ago
The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. what percentage of the mechanical energy of th
Zanzabum

E = ½KA^2 is the mechanical energy of any oscillator.  It is the sum of elastic potential energy and kinetic energy.  When amplitude A decreases by 3%, then

(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)

= {(A2^2 – A1^2) / (A1^2)}

= 97^2 – 100^2/100^2

= 5.91% of the mechanical energy is lost each cycle.

3 0
1 year ago
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