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gayaneshka [121]
3 years ago
5

A 0.1 kg beach ball hit me moving forward at 4 m/s. After hitting me, it bounced back moving at the same speed

Physics
1 answer:
xenn [34]3 years ago
7 0

0.4 N-s is the "impulse" acted on the "beach ball".

Option: C

Explanation:

Given that,

Mass of the "beach ball" is 0.1 kg.

The speed of the ball hits is 4 m/s.

We know that,

Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".

Impulse acted on the beach ball is "mass" times "velocity".

Impulse = mass × velocity

Impulse = 0.1 × 4

Impulse = 0.4 kg m/s

Impulse = 0.4 N-s

Therefore, the "impulse" acted on the ball is 0.4 N-s.

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What is the acceleration of a 19 kg bike pushed with a force of 340 N? F = ma
strojnjashka [21]

Answer:

C

Explanation:

7 0
3 years ago
A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a
Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

T=2\pi \sqrt{\frac{m}{k} }

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0
11 months ago
A basketball player of height 2.40 m is standing 2.60 m in front of a convex spherical mirror of radius of curvature 4.00 m. Det
IrinaVladis [17]

Answer:

The size of the image is 1.04 m.

Explanation:

Given that,

Height of object = 2.40 m

Distance of object = 2.60 m

Radius of curvature =4.00 m

Focal length f=\dfrac{R}{2}=\dfrac{4.00}{2}=2.00

We need to calculate the image distance

Using mirror formula

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{2.00}+\dfrac{1}{2.60}

\dfrac{1}{v}= \dfrac{23}{26}

v=1.13\ cm

We need to calculate the height of the image

Using formula of magnification

m=\dfrac{h'}{h}=-\dfrac{v}{u}

Put the value into the formula

\dfrac{h'}{2.40}=-\dfrac{1.13}{-2.60}

h'=\dfrac{1.13}{2.60}\times2.40

h'=1.04\ m

Hence, The size of the image is 1.04 m

8 0
3 years ago
Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.
nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

  1. The main Ionic bond between the HCOO^- radical and Na^+.
  2. The  sigma covalent bonds between atom of H, atom of C and  both atoms of O.
  3. The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0
3 years ago
If size of an image is 10 cm and the magnification produced by a mirror is +1, what will be the size of object?
olya-2409 [2.1K]

Answer:

h = 10 cm

Explanation:

The magnification produced by a mirror is given by :

m=\dfrac{h'}{h}

h' is the image size and h is the object size

We have, h = 10 cm, m = +1

So,

m=\dfrac{h'}{h}\\\\h=\dfrac{10}{1}\\\\h=10\ cm

So, the size off the object is same as that of the size of the image i.e. 10 cm.

3 0
3 years ago
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