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drek231 [11]
3 years ago
8

For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution

went from zero to 26.8 m/s (60 mi/h) in 3.323 s. Find the magnitude of the car's acceleration.
Physics
1 answer:
Ann [662]3 years ago
3 0

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}

<u>a = 8.06 m/s²</u>

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Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o
Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

         qA = ½ q

         qB = ½ q

         qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

         qA = 1/2 (q / 2) = ¼ q

         qC = ¼ q

         qB = ½ q

At this point all spheres have a charge,

      qA = ¼ q

      qb = ½ q

      qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

       Q = ½ q + ¼ q = ¾ q

When splitting the charge

        qB = ½ ¾ q = 3/8 q

        qC = ½ ¾ q = 3/8 q

        qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

       Q = qA + QB + QC

       Q = ½ q + 3/8 q + ¼ q

        Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0
3 years ago
Need help ASAP
devlian [24]

Answer:

cell pohnes convert sound waves into radio waves

electromagnetic waves used in cellphone communications are called: radio waves

To send out a radio signal far and wide.. it is called: broadcast

Explanation:

5 0
3 years ago
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A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
ollegr [7]
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
8 0
3 years ago
A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

3 0
3 years ago
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Explain why a steel block sinks but a steel ship floats
uysha [10]

Answer:

Fluids exert forces on objects because of many molecules of the fluid that continuously collide with the surfaces of the object immersed in the fluid. ... A steel boat floats on water but a steel block does not because the block has more weight than the steel boat due to the buoyant force.

Explanation:

3 0
3 years ago
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