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Rzqust [24]
2 years ago
14

Please help asap!!!!!!!!!!!

Physics
1 answer:
Tomtit [17]2 years ago
8 0
The answer would be 20000

The answer in standard form/scientific notation would be 2 x 10^4

(The exponent is 4 because that's how many digits after the 2 there is)
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Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.
Snezhnost [94]

 

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

 

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

 

Therefore the train has a constant acceleration of about 0.23 meters per second.

5 0
4 years ago
What is the net force needed to lift a full grocery sack weighing 210N uniformly?
Alex
If the sack weighs 210 newtons, then an upward force of 210 newtons
exactly cancels the downward force of gravity, and makes the net vertical
force on the bag zero. 

ANY upward force that's greater than 210 newtons makes the net force
act upward on the bag, and causes it to accelerate upward.
8 0
3 years ago
Newspapers used as covers to keep out the cold were called _________.
Nata [24]
The answer is hoover blankets♥️
5 0
4 years ago
A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 475°C. The aluminum plate has
Anuta_ua [19.1K]

Answer:

T_0 = 338.916 Degree\ celcius

Explanation:

Given data:

Thickness of aluminium sheet 10 cm

initial temperature = 25 degree celcius

Assumption

Thermal properties remain constant, transfer of heat by radiation is negligible.

from the information given in the question we have

T_S ≈T_∞ , it implies we have h → ∞

from table 4.2 Biot number → ∞ the value of

\lambda_1 = 1.5708 and A_1= 1.2732

The fourier number is

t = \frac{\alpha t}{l^2} = \frac{97.1\times 10^{-6} \times 15}{0.05^2} = 0.5826

Temperature at center after 15 second of heating

\theta _{0 wall} = \frac{T_0 - T_{\infity}}{T_i -T_{\infity}} = A_i e^{\lambda_1^2 t}

T_0 = T_i -T_{\infity} \times A_i e^{\lambda_1^2 t}

T_0 = (25 - 475) 1.2732 e^{-1.5708^2 \times 0.5826} +  475  = 356 degree celcius

T_0 = 338.916 Degree\ celcius

8 0
3 years ago
A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t
Nataly [62]

Answer:

Explanation:

Given

N0 = 20kg (original substance)

decay constant λ = 0.179/sec

time t = 300s

We are to find N(t)

Using the formula;

n(t) = N0e^-λt

Substitute the given values

N(t) = 20e^-(0.179)(300)

N(t) = 20e^(-53.7)

N(t) = 20(4.7885)

N(t) =143.055

To know how much of the original material that is active, we will find N(t)/N0 = 143.055/20 = 7.152

About 7 times the original material is still radioactive

4 0
3 years ago
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