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Kazeer [188]
3 years ago
7

How does speed relate to the distance covered and the time taken for travel?

Physics
1 answer:
Tcecarenko [31]3 years ago
7 0

Think MPH on a car, The Faster you go the less time it takes to get there. For example, the car again, if a red and blue car are both start in Arizona and want to  go to Disney Land they leave the same at the same time but the Blue car goes 50 mph and the Red car 75 mph. The Red car will get there first.


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calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin
Lynna [10]

Answer:

minimum number of photon is 4.05 × 10^{7}

Explanation:

given data

energy = 50 keV = 50 × 10^{3} eV =  50 × 10^{3} × 1.602× 10^{-19}  J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E  = n  × h  × ν   .......................1

put here all these value

50 × 10^{3} = n × 6.6× 10^{-34} × c/ 1× 10^{-3}

50 × 10^{3} × 1.602× 10^{-19}  = n × 6.6× 10^{-34} ×( 3 × 10^{8} / 1× 10^{-3})

solve it and find n

n = 4.05 × 10^{7}

so here minimum number of photon is 4.05 × 10^{7}

3 0
3 years ago
What is transmitted by all waves? energy mass matter sound
disa [49]

Answer: energy

Explanation: A wave is a disturbance that transfers energy as it travels through medium.

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3 years ago
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saw5 [17]

Explanation:

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3 years ago
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Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha
Ann [662]

Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

8 0
2 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
2 years ago
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