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3 years ago

The energy required to completely remove the covalent bond between two

Physics

1 answer:

hoa [83]3 years ago

7 0

Answer:

B is the answer. Correct me if I'm wrong

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A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

trasher [3.6K]

Answer:

$\boxed {\boxed {\sf 500 \ Newtons }}$

Explanation:

The equation for force is given:

$F=m*a$

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

$m= 100 \ kg \\a= 5 \ m/s^2$

Substitute the values into the formula.

$F=100 \ kg * 5 \ m/s^2$

Multiply.

$F= 500 \ kg*m/s^2$

1 kilograms meter per square second is equal to 1 Newton.
Our answer of 500 kg*m/s² is equal to 500 Newtons.

$F= 500 \ N$

The force exerted by the go cart engine is 500 Newtons.

4 0

2 years ago

Read 2 more answers

First right is brainliest, plz help:)

Aleonysh [2.5K]

They are magnetic

Your welcome;)

5 0

2 years ago

Reflecting telescopes are popular because they're

musickatia [10]

The answer is letter C.

Reflecting telescopes are more powerful than refracting telescopes. These are also called as reflectors which serves an optical telescope that uses a single or combination of curved mirrors. These mirrors then reflect light and form an image. It is designed for very large diameter objects and are mostly considered as major telescopes in the field of astronomy. They were used as an alternative for refracting telescopes during the 17th century because they suffer less chromatic aberrations than a refracting telescope does.

7 0

2 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

2 years ago

In order to increase the amount of work done, we need to:

Nadusha1986 [10]

Option (D) is the correct one.

In order to increase the amount of work done, we need to increase the force applied to the object.

8 0

3 years ago