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bulgar [2K]
2 years ago
10

A car accelerates from rest to 90m/s in 3 seconds. What is the acceleration and how far did it travel?

Physics
1 answer:
Inessa [10]2 years ago
4 0

Answer:

Explanation:

1.)acceleration= final velocity-initial velocity/time

90m/s-0m/s /3seconds

=30m/s^2

2.) distance= speed*time

90m/s*3seconds

=270m

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How does the sun create heat?
Triss [41]

Explanation:

sun was created through the hypothesis

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2 years ago
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An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A
Snezhnost [94]

Answer:

9.96x10^-20 kg-m/s

Explanation:

Momentum p is the product of mass and velocity, i.e

P = mv

Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpha radioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light.

From this we calculate the speed as

v = 5% 0f 3x10^8 m/s (speed of light)

v = 1.5x10^7 m/s

The mass of an alpha particle is approximately 6.64×10−27 kg

Therefore,

P = 1.5x10^7 x 6.64×10^−27

P = 9.96x10^-20 kg-m/s

8 0
2 years ago
When popcorn is heated over a flame, the kernel burst open. why does this occur?
Bezzdna [24]
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4 0
3 years ago
What is density?
il63 [147K]
<span>a. the amount of matter in a given volume </span>
8 0
2 years ago
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A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
2 years ago
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