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dlinn [17]
2 years ago
8

What constant acceleration is required to increase the speed of a car from 27 mi/h to 53 mi/h in 3s?

Physics
1 answer:
mario62 [17]2 years ago
7 0

Answer:

8.87 m/s^2  

Explanation:

Acceleration is change in velocity over a change in time

27 to 53 = 26 m/s   change in velocity

26 m/s / 3 s = 8.87 m /s^2

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You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti
vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

5 0
3 years ago
Identify the statement below that is true about a type of stress.
Sunny_sXe [5.5K]

1-D Eustress is a positive response

2-B resistance

3-A Imagine you are doing your gymnastics

6 0
3 years ago
Read 2 more answers
A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. immediately after the impact
pshichka [43]
Refer to the diagram shown below.

Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.

Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.

Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s

100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s

V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°

Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s


8 0
3 years ago
Read 2 more answers
Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol
Tju [1.3M]

Answer:

R \approx 2.418\times 10^{9}\,km

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

T = \frac{2\pi}{\omega}

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

\omega = \sqrt{\frac{a_{r}}{R} }

Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}

Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

R \approx 2.418\times 10^{9}\,km

4 0
4 years ago
Distance is a vector quantity ... true or false
Elena-2011 [213]
Answer:False
Reason:
7 0
3 years ago
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