A machine does 1200 J of work in 1 min. What is the power developed

A student has designed a circuit with one battery, two light bulbs, and two switches, pictured to the left. She would like for e

Gnoma [55]

Answer:

The bulb is not powered because the negative end of the battery is not connected to the electric current,making it dysfunctional. The electrons have to flow through the positive and negative ends of the battery in order for it to work.

Hope this helps.

5 0

3 years ago

Read 2 more answers

A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?

vovikov84 [41]

We can solve the problem by using Newton's second law of motion:
$F=ma$
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
$a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2$

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>

5 0

4 years ago

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista

Nastasia [14]

Answer:A)u =4.295m/s , B)a = 29.746m/s² C) F=3,153N

Explanation:

Using the kinematic expression

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity = 0

Here, acceleration due to gravity is acting in opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening = 4.295m/s,

a = acceleration while straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as

a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

F = 3,153.076 rounded to 3,153N

8 0

3 years ago

A 1000-kilogram car traveling due east at 15 meters per second hit from behind and receives a forward impulse of 6000 newton-sec

strojnjashka [21]

The change in momentum of the car is 6000 kg m/s

Explanation:

According to the impulse theorem, the change in momentum of an object is equal to the impulse exerted on the object, therefore:

$\Delta p = I$

where

$\Delta p$ is the change in momentum

I is the impulse exerted

For the car in this problem, the impulse received is

I = 6000 kg m/s (in the forward direction)

Therefore, the change in momentum of the car is equal to this value:

$\Delta p = I = 6000 kg m/s$ (in the forward direction)

We can also calculate what is the new momentum of the car. In fact, the initial momentum is

$p_i = mu = (1000 kg)(15 m/s)=15,000 kg m/s$

And so, the new momentum is

$p_f = p_i + \Delta p = 15,000 + 6,000 = 21,000 kg m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

8 0

3 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$