What is this?<br> Picture

A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.

Vaselesa [24]

V2=u2+2as
v2=144+600
v2=744
v=√744

6 0

3 years ago

2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to

ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b) the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

F = m a

a = F / m

Let's use kinematics to find the velocity after x = 1 m

v² = v₀² + 2 a x

The initial speed is zero

v = √ (2 F/m x)

For the distance of x = 1 m

v₁ = √ (2 F / m)

Car 2

F = 2m a

a = F / 2m

v² = 2 a x

v = √ (F/m x)

For x = 1 m

v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m 2F / m

K₁ = F

Car 2

K₂ = ½ 2m v₂²

K₂ = ½ 2m F / m

K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m √ (2F / m)

Car 2

P₂ = 2m v²

P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

v = v₀ + a t

v = F / m t

v₁ = F / m 10

Car 2

v₂ = F / 2m 10

v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m (F / m 10)²

K₁ = 50 F² / m

Car2

K₂ = ½ 2m v₂²

K₂ = m (F / m 5)²

K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m F / m 10

P₁ = 10 F

Car 2

P₂ = 2m v₂

P₂ = 2m F / m 5

P₂ = 10 F

In this case the moment of the two cars is the same

7 0

3 years ago

When observing a specimen in a microscope you are told that the total magnification of the specimen is 630x assuming you are usi

Norma-Jean [14]

<span>Objective Lenses: Usually you will find 3 or 4 objective lenses on a microscope. They almost always consist of 4X, 10X, 40X and 100X powers. When coupled with a10X (most common) eyepiece lens, we get total magnifications of 40X (4X times10X), 100X , 400X and 1000X.</span>

5 0

3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

So

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

7 0

3 years ago

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Ainat [17]

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

$\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg$