Ask question

Ask question

All categories

3 years ago

13

As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre

ases proportionally D. decreases proportionally

2 answers:

diamong [38]3 years ago

4 0

Answer:

USE SOCRACTIC IT WOULD REALLY HELP

Aleksandr-060686 [28]3 years ago

3 0

The answer is c. as an object is in motion speeds up or “becomes heavier”, it’s kinetic energy increases proportionally: double the velocity and you quadruple the kinetic energy. this is why a tiny bullet traveling at high speed does so much more damage than a huge truck bumping into something at 1 mph. so the answer is c

You might be interested in

an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni

stiv31 [10]

The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N

3 0

2 years ago

Pls Help Me with this problem

Naddika [18.5K]

Answer:

explanation

Explanation:

1 = C

2 = A

3 = D

4 = E

5 = B

8 0

2 years ago

Read 2 more answers

How does a vacuum flask keep drinks hot? Explain in terms of conduction, convection and radiation.

Flura [38]

Vacum flask has inner layer of glass and outer body of plastic.the glass has a reflective metal layer..when a hot drink is poured into the flask it prevents conduction.tight lid stops the convection.when radiation tries to leave the hot drink the reflective layer of the glass reflects it back so no heat can escape.

4 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o

defon

Part a)

Magnitude of electric field is given by force per unit charge

$E = \frac{F}{q}$

$E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}$

$E = 2150 N/C$

Part b)

Electrostatic force on the proton is given as

F = qE

$F = 1.6 * 10^{-19} * 2150$

$F = 3.44 * 10^{-16} N$

PART C)

Gravitational force is given by

$F_g = mg$

$F_g = 1.6 * 10^{-27}*9.8$

$F_g = 1.57 * 10^{-26} N$

PART d)

Ratio of electric force to weight

$\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}$

$\frac{F_e}{F_g} = 2.2 * 10^{10}$

7 0

3 years ago