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Maksim231197 [3]
1 year ago
10

A change in the speed or direction of an object is called

Physics
2 answers:
tangare [24]1 year ago
8 0

Answer:

Hey Dude,

Explanation:

This is ur answer...

<h2><em>Acceleration is any change in the speed or the direction of an object's motion.</em></h2>

Hope it helps!

Brainliest pls!

Follow me!

neonofarm [45]1 year ago
5 0

Answer:

Acceleration

Explanation:

Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.

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A bullet of mass 0.01 kg is fired in to a sand bas of mass 0.49 kg from a tree. The Sand bag with the bullet embedded in to it s
makvit [3.9K]

Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.

<h3>Linear momentum</h3>

The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.

Momentum before collision for the bullet = 0.01 kg × v

Momentum before collision for the bag = 0

Momentum after collision for the bag and bullet = (0.01 kg  +  0.49 kg) 10 = 5 Kgms-1

The velocity of the bullet before collision =  0.01 kg × v + 0 =  5 Kgms-1

v = 5 Kgms-1/0.01 kg

v = 500 ms-1

Learn more about momentum: brainly.com/question/904448

8 0
2 years ago
Read 2 more answers
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

5 0
3 years ago
Whats it called when a force can change the shape of an object?
alekssr [168]

Force applied causes deformation in the object. It changes the relative positions of constituent particles in the crystal lattice.
As soon as that happens, the interatomic or intermolecular forces come into play and they, tend to restore the solid back to it's original shape.

This restoring force per unit area is called Stress . When external forces are removed, the internal forces tend to restore the solid back.
This property is called Elasticity .

However, no material is perfectly elastic and what happens is that, the body is not able to restore itself completely.

 
3 0
3 years ago
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our 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between th
gtnhenbr [62]

Answer:

Explanation:

Maximum force of friction possible = μmg

= .65 x 3.8 x 9.8

= 24.2 N

u = 72 x 1000 / 60 x 60

= 20 m /s

v² = u² - 2as

a = 20 x 20 / (2 x 30)

= 6.67 m / s²

force acting on it

= 3.8 x 6.67

= 25.346 N

Friction force possible is less .

So friction will not be able to prevent its slippage

It will slip off .

4 0
3 years ago
Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is
Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = \frac{K\times Q}{d^2}

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically  opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two  charges will be zero.  

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0
3 years ago
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