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3 years ago

If we connect a third bulb in our series circuit, say one with 4

Physics

1 answer:

Alexus [3.1K]3 years ago

6 0

Answer:

The current will decrease.

Explanation:

When another bulb is added, the resistance is going to increase. Keep in mind that the current is inversely proportional to the resistance (Ohm's law: R= V/I ). Therefore when the resistance increase, the current running in the circuit will decrease.

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago

Nuclear _________ reactions convert protons into helium; thus, becoming the source of all energy radiated by the sun. A) fission

Bogdan [553]

Nuclear fusion reactions convert protons into helium; thus, becoming the source of all energy radiated by the sun.

Answer: Option D

Explanation:

Nuclear fusion reaction is one among the two nuclear reactions in which the atoms nucleus interact with each other to produce the products. In nuclear fusion, two smaller atoms react together to form a new atom with bigger size.

So large amount of energy is required to start the nuclear fusion reaction. The fusion reactions mostly occurs in stars. The illumination in Stars even the Sun is due to nuclear fusion reaction occurring with the atoms present in them.

Mostly the energy radiated by Sun is due to the protons-protons chain reaction. In this chain reaction, the protons get converted into helium due to nuclear fusion reaction, thus becoming the source of all energy radiated by the Sun.

3 0

3 years ago

Anna Litical analyzes the force between a planet and its moon, varying the mass of

Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

Trial 1

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

Trial 2

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

Trial 3

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)

Trial 1 is the largest, trial 3 is the smallest

5 0

1 year ago

Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv

ehidna [41]

The best option is B) 7.0 × 10² newtons.

If Earth attracts a person with a gravitational force of 7.0 × 10² newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

5 0

3 years ago