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Svetllana [295]
3 years ago
5

if a car has a constant acceleration of 4 m/s^2, starting from rest, how fast is it traveling after 5 seconds?​

Physics
2 answers:
UkoKoshka [18]3 years ago
3 0

Answer:

20 m/s

Explanation:

Recall that one of the equations of motions can be written:

v = u + at, (also see attached for reference)

Where,

v = final speed (we are asked to find this)

u = initial speed = 0 (because it starts from rest)

t = time taken = 5s

We simply substitute the given values into the equation:

v = u + at

v = 0 + (4)(5)

v = 20 m/s

Anni [7]3 years ago
3 0

Answer:If a car has a constant acceleration of 4 m/s2, starting from rest, how far has it traveled after 5 seconds? The distance travelled in 5s.

Explanation:

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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s
ozzi
The answer is A- feeling a !etal wire get warmer as you roast a !arshmallow over a fire
3 0
2 years ago
Read 2 more answers
Which element is a metalloid?
Ann [662]

Answer:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

All of the possible metalloids are:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)

3 0
1 year ago
Two sealed 1 l containers full of gas are at room temperature. Container a has a pressure of 4 atm and container b has a pressur
loris [4]

Answer:

The number density of the gas in container A is twice the number density of the gas in container B.

Explanation:

Here we have

P·V =n·R·T

n = P·V/(RT)

Therefore since V₁ = V₂ and T₁ = T₂

n₁ = P₁V₁/(RT₁)

n₂ = P₂V₂/(RT₂)

P₁ = 4 atm

P₂ = 2 atm

n₁ = 4V₁/(RT₁)

n₂ =2·V₁/(RT₁)

∴ n₁ = 2 × n₂

Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.

This can be shown based on the fact that the pressure  of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.

5 0
3 years ago
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