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3 years ago

How much faster will a car be going if it accelerates at 1 m/s/s for a total of 60 seconds.

Physics

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer:

Vf = 60 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}= v_{i}+(a*t)$

where:

Vf = final velocity [m/s]

Vi = initial velocity = 0

a = acceleration = 1 [m/s^2]

t = time = 60 [m]

Now replacing:

Vf = 0 + (1*60)

Vf = 60 [m/s]

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a car increases its speed as it moves across the floor. which form of energy is increasing for the car?

zavuch27 [327]

Answer:

kinetic

Explanation:

i just remember it from last year

6 0

2 years ago

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.

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3 years ago

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I am looking to buy a new car this summer. I want to buy the fastest car I can and in my

Mashcka [7]

Answer

Se togli 15 mph da 95 e 15, capisci quanto tempo la macchina 2 fa da 0 mph a 70 mph. La prima macchina fa da 0 mph a 60 mph in 5 secondi, e la seconda da 0 mph a 70 mph in 5 secondi. Risulta essere più veloce la seconda macchina. Spero di essere stato utile :)

Explanation:

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2 years ago

NemiM [27]

Answer:

what is your question because I seem to not see a question in this question....

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2 years ago