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Anna71 [15]
3 years ago
7

Best exercises for these topics. (have the amount of exercises, have 4 exercises for each.)

Physics
2 answers:
Kipish [7]3 years ago
5 0

Answer:

what is the question?

Explanation:

i dont understand

yaroslaw [1]3 years ago
5 0

Answer:

follow this routine .

Explanation:

because it can build up your body fast and effectively so

You might be interested in
(50 Points)
Sauron [17]

Answer:   The piezoelectric effect describes the property of certain solids to generate an electrical charge under mechanical stress. The actuating force modifies the microscopic structure of the body; dipoles are created between which a voltage builds up. And Their positive and negative charges complement each other creating two types of electricity: static electricity

Explanation:

6 0
2 years ago
Mary is driving a car of mass 900 kilograms toward the north with a velocity of
lorasvet [3.4K]
Since both cars move together after the collision, then this is an example of an inelastic collision. The formula for an inelastic collision is as follows:

m1u1 + m2u2 = (m1 + m2)v

Where:

m1 = mass of the first object
m2 = mass of the second object
u1 = initial velocity of the first object
u2 = initial velocity of the second object
v = final velocity

Substituting the given values to solve for v:

900*22 + 900*15 = (900 + 900)v
v = 18.5 m/s
3 0
4 years ago
Read 2 more answers
Una cámara fotográfica analógica (no digital) tiene dos lentes intercambiables. Uno de foco 55mm y el otro de 200 mm. Toma una f
Sergeu [11.5K]

Answer:

f = 55mm,     h ’= -9.89 cm

f = 200 mm,  h ’= 42.5 cm

Explanation:

For this exercise let's start by finding the distance to the image, using the equation of the constructor

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and image, respectively

lens with f₁ = 55mm = 0.55cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{0.55} - \frac{1}{10}

          \frac{1}{q_1} = 1.718

          q₁ = 0.582 m

lens with f₂ = 200mm = 2m

           \frac{1}{q_2} =   \frac{1}{2} - \frac{1}{10}

            \frac{1}{q_2} = 0.4

            q₂ = 2.5 m

the magnification of a lens is given by

            m = \frac{h'}{h} = -  \frac{q}{p}

             h ’= - \frac{q}{p} \ h

let's calculate for each lens

f = 55mm

             h '= - 0.582 / 10 1.7

             h ’= 0.0989 m

             h ’= -9.89 cm

f = 200 mm

             h '= - 2.5 / 10 1.7

             h ’= -0.425 m

             h ’= 42.5 cm

The negative sign indicates that the image is real and inverted

4 0
3 years ago
A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t
vladimir1956 [14]
Mass of yellow train, my = 100 kg

Initial Velocity of yellow train, = 8 m/s

mass of orange train = 200 kg

Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)

To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

The sum of initial momentum = the sum of final momentum


Since the question only wants the sum of initial momentum,

(100)(8) + (200)(-1) = 600 m/s

8 0
2 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time
Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

 

4 0
4 years ago
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