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soldier1979 [14.2K]
2 years ago
6

An 8.2 kg object accelerates at 8.0 m/s^2. What is the acceleration?

Physics
1 answer:
ZanzabumX [31]2 years ago
6 0

Answer:

8.0m/s²

Explanation:

meters per second squared is a unit of acceleration so 8.0m/s² is the answer

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A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6
Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0
3 years ago
What will happen to an object's wavelength as the object moves towards you?
Ilia_Sergeevich [38]
The wavelength will get shorter
5 0
3 years ago
Read 2 more answers
If a person looking at a poster sees green instead of yellow and doesn't see red at all, this person most likely has color blind
katrin [286]
Red - sensitive so answer is c
7 0
3 years ago
Read 2 more answers
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at
klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

\lambda = Wavelength

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

Energy of a photon is given by

E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}

Let \lambda_1 = 700 nm

\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}

For red light

E_1=\dfrac{hc}{\lambda_1}

For UV light

E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}

Dividing the equations

\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1

Hence, the answer is  b) twice the energy of each photon of the red light.

7 0
3 years ago
Read 2 more answers
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