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natka813 [3]
3 years ago
10

Firecrackers 1 and 2 are 600 m apart. You are standing exactly halfway between them. Bob is standing 300 m on the other side of

firecracker 1. You see the flashes of light from the two firecrackers exploding at the exact same instant.
a) Does Bob see both firecrackers explode at exactly the same time, firecracker 1 explode first, or firecracker 2 explode first? Explain.
Physics
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

the same time

Explanation:

since the speed of light is very fast (3×10^8 ms^-1) the difference in time arrived from firecrackers A and B to Bob is negligible, and our brain can distinguish such small time difference,l.

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
VLD [36.1K]

Answer:

(a)  8.2 x 10^-8 N

(b) 3.6 x 10^-47 N  , 2.27 x 10^39

Explanation:

charge of proton, q1 = 1.6 x 10^-19 C

charge of electron, q2 = - 1.6 x 10^-19 C

radius of orbit, r = 0.053 nm = 0.053 x 10^-9 m

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

Gravitational constant, G = 6.67 x 10^-11 Nm^2/kg^2

(a) The electrostatic force between two charges is given by

F_{e}=\frac{Kq_{1}q_{2}}{r^2}

Where, K is the coulombic constant = 9 x 10^9 Nm^2/C^2

By substituting the values

F_{e}=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\1.6\times 10^{-19}}{\left ( 0.053 \times 10^{-9} \right )^2}

Fe = 8.2 x 10^-8 N

(b) The gravitational force between the electron and proton is given by

F_{g}=\frac{Gm_{e}m_{p}}{r^{2}}

F_{g}=\frac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\1.67\times 10^{-27}}{\left ( 0.053 \times 10^{-9} \right )^2}

Fg = 3.6 x 10^-47 N

\frac{F_{e}}{F_{g}}=\frac{8.2\times10^{-8}}{3.6\times 10^{-47}}

\frac{F_{e}}{F_{g}}=2.27\times 10^{39}

5 0
3 years ago
how does a sprinter sprint what is the forward force on a sprinter as she accelerates. where does that force come from
NARA [144]

On the starting blocks, sprinters use their feet to push backward. The blocks respond by pressing forward with a force equal to this with their feet.

<h3>What drives the sprinter forward?</h3>

Vertical forces must be larger than the pull of gravity in order to assist the sprinter in moving forward as gravity is pushing him or her downward. The propulsive force is the force that propelled the runner forward.

<h3>Basketball players must jump straight up into the air, but how?</h3>

An interaction diagram and a free-body diagram should be included in your explanation. The player pushes down on the ground, which pushes up against him in return. As a result of his push being stronger than gravity, the player accelerates upward.

To know more about sprinter forward visit:-

brainly.com/question/14310782

#SPJ4

8 0
1 year ago
A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container
Len [333]

Answer:

This is a heat balance question with negligible heat losses  to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

3 0
4 years ago
When a force is applied to an object, no work is being preformed on the object unless the object____.
Temka [501]
No work is being performed on the object unless the object is moving, as work done = force x distance
8 0
3 years ago
Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107m above Earth's surface (that is, if it were three Earth ra
Vladimir79 [104]
The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N
5 0
3 years ago
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