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2 years ago

What is the kinetic energy of a soccer ball which has a mass of 0.8 kg and is kicked at a velocity of 10 m/s

Physics

1 answer:

Lubov Fominskaja [6]2 years ago

5 0

Answer:

Explanation:

Kinetic energy= K.E = (1/2) × mv²

K.E = 0.5× 0.8kg× 100m²/s² = 40 N

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Two identical metal balls are rolling without slipping along a horizontal surface with speed V. Each ball encounters a hill ('tw

Anastaziya [24]

Answer:

The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.

Explanation:

Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.

The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.

Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill

3 0

2 years ago

A stock person at the local grocery store has a job consisting of the following five segments:

vaieri [72.5K]

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

6 0

2 years ago

A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold

salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0

2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

2 years ago