The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?

Answers

Answer 1

Answer:

Final answer:

The problem discusses the change in acceleration when a passenger is added to a car. It requires understanding of Newton's second law of motion, force equals mass times acceleration, and then recalculating the acceleration with the passenger added to the total mass.

Explanation:

This problem pertains to Newton's second law of motion, stating that the force applied on an object equals its mass times its acceleration (F = ma). Given that the initial acceleration of the Lamborghini Huracan with a driver is 0.80g or 0.80*9.80 m/s², we can calculate the force applied by the car. By multiplying the car's mass (1510 kg) with its acceleration, we will find the force.

Οnce we have the force, we can calculate the new acceleration if the 80 kg passenger rode along. Given that the force is constant, we determine the car's new acceleration by dividing this force with the new total mass (car mass + passenger's mass). So the question ultimately requires an application of the concepts of force, mass, and acceleration.

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Answer 2

Answer:

Final answer:

The new acceleration of the Lamborghini Huracan with an added passenger can be calculated by finding the initial force using the car's mass and acceleration, and then using this force with the increased mass (original mass + passenger's mass) to find the new acceleration. The new acceleration will be less than the initial acceleration due to the increased mass.

Explanation:

To determine the new acceleration of the Lamborghini Huracan with an added passenger, we first calculate the initial force acting on the car. This can be done by using Newton's second law which states that Force = mass * acceleration. Initially, the acceleration is 0.80g (where g is acceleration due to gravity = 9.81 m/s²), and the mass is 1510 kg (including the driver). Therefore, the initial force = 1510 kg * 0.8 * 9.81 m/s².

However, when an 80-kg passenger rides along, the total mass becomes 1510 kg + 80 kg = 1590 kg. To find the new acceleration, we keep the force constant (as it is not affected by the introduction of the passenger) and rearrange the formula F = m*a as a = F/m. Use the increased mass to find the new acceleration. Please note that the new acceleration will be less than the initial acceleration due to increased mass.

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Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

Consider a proton travelling due west at a velocity of 5×10^5m/s. Assuming that the rth magnetic field has a strength of 5x10^-5Tand is directed due south calculate li) the magnitude of the force on the proton (q= 1.6x10^-9C)

Answers

Answer:

Explanation:

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.89 Fcapacitor charged to 60.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 431, with one possessing a dielectric constant of 947.Required:
a. Find the electric potential energy of the original capacitor when it is charged. (in Joules)
b. Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

Answers

Answer:

$U = 3.505 *10^9 \ J$

$U_1 = 7.696 *10^9 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 1.89 \ F$

The voltage is $V = 60.9 \ k V = 60.9 *10^(3) \ V$

The first dielectric constant is $\epsilon_1 = 431$

The second dielectric constant is $\epsilon_2 = 947$

Generally the electric potential energy is mathematically represented as

$U = (1)/(2) * C * V^2$

=> $U = (1)/(2) * 1.89 * (60.9 *10^(3))^2$

=> $U = 3.505 *10^9 \ J$

Generally the capacitance when the capacitor's filling was changed is

$C_n = 1.89 * (947)/(431)$

=> $C_n = 4.15$

Generally the electric potential energy when the capacitor's filling was changed is

$U_1 = (1)/(2) * C_1 * V^2$

=> $U_1 = (1)/(2) * 4.15 * (60.9 *10^(3))^2$

=> $U_1 = 7.696 *10^9 \ J$

If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mix? Explain your answer.

Answers

Answer:

Mix

Explanation:

A battery has two electrodes at both of its end terminals namely the anode which is the negatively charged electrode and the anode which is the positively charged electrode.

Now, Electrons usually travel through the battery circuit from the anode to the cathode, and this motion of travel is the propelling force that makes electricity flow through the circuit.

Now, the anode will need to have a low electron affinity because it needs to easily release electrons during discharge while the cathode needs to have a high electron affinity because it normally readily accept electrons during discharge.

Thus, for a battery to be strong, it is a combination of high electron affinity and low electron affinity.

Answer: mix

Explanation:

you cant pair 2 of the same electron affinities

1 microgram equals how many milligrams?

Answers

Answer: 1 microgram is equal to 0.001 miligrams

Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?