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bazaltina [42]
3 years ago
11

John is preparing for his way to school. His house is 30 km away. What time will he be in school if he leaves his house at 6:30

AM and travels at an average speed of 40 km/hr?
Physics
1 answer:
Goshia [24]3 years ago
7 0

Answer:

Jio will be at school at 7:15 a.m.

Explanation:

Formula:

t= s/v =(30km)/(40km/h) =0.75•60min=45min.

Then, Jio will be at school at 6:30a.m. + 45min =7:15a.m.

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Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a
Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

h = \frac{gt^2}{2}

solving for t

we have

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

substituing all value for time t

t = \sqrt{\frac{2\times 11.98}{9.8}}

t = 1.56 s

we know that speed is given as

v = \frac{d}{t}

v =\frac{11.98}{1.56}

v = 7.67 m/s

7 0
3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

5 0
3 years ago
If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c
Sloan [31]

Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2  into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

q = 4.87e-14 C

6 0
3 years ago
Sound waves move through some liquid at a speed of 1500 m/s. If the wavelength is 3 m, what is the frequency of the wave?
Orlov [11]

Answer:500hz

Explanation:

Frequency=velocity/wavelength

Frequency=1500/3

Frequency=500hz

4 0
3 years ago
What type of motion does an object have if it is falling with terminal velocity ?
Arte-miy333 [17]

Answer:

I believe force :|

Explanation:

4 0
3 years ago
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