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3 years ago

A 1000-kg car has 50,000 joules of kinetic energy. What is the speed?

Physics

2 answers:

lesantik [10]3 years ago

5 0

Answer:

½mv²= KE

½(1000)?²= 50000

500 500

? ²=100

Square root

?= 10

10 m/s is you answer

Arlecino [84]3 years ago

3 0

Kinetic Energy = 1/2×m×v²
50000 = 1/2×1000×v²
v = √100
v = 10 m/s

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an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo

Otrada [13]

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

$F = ma$

$9000 = 6000 a$

$a = 1.5 m/s^2$

now for finding minimum stopping distance of the truck

$v_f^2 - v_i^2 = 2a d$

$0^2 - 15^2 = 2(-1.5) d$

$d = 75 m$

4 0

3 years ago

A box experiences a force of 2 N to the left and 3 N to the right. Which is true of the box's motion?

Nostrana [21]

Below are the choices:

<span> A) The box will slow down.
B) The box's velocity will be 1 m/s.
C) The box's velocity will not change.
D) The box will experience acceleration
</span>
The answer is D) The box will experience acceleration

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5 0

3 years ago

Read 2 more answers

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :