3 years ago

Big Bee has a mass of 5 kg and hits a newly cleaned windshield accelerating at a rate of 2 m/s2.

Physics

1 answer:

Juli2301 [7.4K]3 years ago

8 0

Answer

The Little Bee will have a greater force of 12N

<u>Explanation</u>

As we know that Force is equal to the product of mass and acceleration (F=ma)

The Force with which the Big Bee will hit the windshield = (5kg x 2m/s^2)

Force of Big Bee= 10 Newton (N)

The Force with which the Little Bee will hit the windshield = (3kg x 4m/s^2)

Force of Little Bee = 12N

Hence, the little bee will hit the windshield with a greater force.

You might be interested in

What do we call a substance in<br> which two or more elements are<br> chemically bonded

Goryan [66]

Answer:

A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

8 0

3 years ago

Read 2 more answers

If the potential difference across the bulb in a camping lantern is 9.0 V, what is the

Gre4nikov [31]

Answer:

Explanation:

The potential difference across the terminal as the same and thats because we are assuming that the source has no internal resistance.

Internal resistance are usually little resistances in the supply.

4 0

2 years ago

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

NeX [460]

Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0

3 years ago

A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base

riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

$h=vt+\dfrac{1}{2}gt^2$

v= 0 m/s

$h=\dfrac{1}{2}gt^2$

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0

2 years ago

A small 25 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe’s kinetic energy?

Helen [10]

Kinetic energy = 1/2mv^2

=1/2(25)(1^2)
= 12.5J

4 0

3 years ago