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3 years ago

Which arrow best represents the path of an object with projectile motion?

Physics

1 answer:

bonufazy [111]3 years ago

6 0

Answer:

Explanation:

As should have been explained at the very start of the lesson, it states "Whatever goes up, must come down."

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A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

baherus [9]

20 characters longer later and woah

5 0

3 years ago

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0

3 years ago

A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back

saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

$K.E =\dfrac{1}{2}mv^2$

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

$K.E=\dfrac{1}{2}\times0.5\times(4)^2$

$K.E = 4\ J$

The average force exerted by the floor on the ball = change in kinetic energy over collision time

$F = \dfrac{4}{0.05}$

$F=80\ N$

Hence, The force exerted by the floor is 80 N.

4 0

3 years ago

A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block

mel-nik [20]

Answer:

v₁ = 0.375 m / s , x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

$p_{f}$ = (m + M) v₁

how the forces are internal

p₀ =p_{f}

m v₀ = (m + M) v₁

v₁ = m /m+M v₀

let's calculate

v₁ = 0.4 /(0.4 + 2.8) 3

v₁ = 0.375 m / s

Let's apply Newton's second law to the Block, to find the friction force

Y axis

N - W = 0

N = W

N = m g

where m is the mass of the block

the friction force has the formula

fr = μ N

fr = μ m g

We apply Newton's second law to slab

X axis

fr = M a

where M is the mass of the slab

μ m g = M a

a = μ g m / M

let's calculate

a = 0.15 9.8 0.4 / 2.8

a = 0.21 m / s²

With kinematics we can find the position

v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

v² = 2 a x

x = v2 / 2a

let's calculate

x = 0.375²/2 0.21

x = 0.335 m

4 0

3 years ago

Two long, parallel wires carry currents of different magnitudes. if the amount of current in each wire is doubled, what happens

Mrac [35]

The force per unit of length between two wires carrying current is
$\frac{F}{L}= \frac{\mu_0 I_1 I_2}{2 \pi r}$
where I1 and I2 are the currents in the two wires, while r is the distance between them.

We can see from the formula that the force is proportional to the product between I1 and I2: $F \sim I_1 I_2$
so, if we double both I1 and I2, we get a factor 4:
$F' \sim (2I_1 )(2I_2)=4 I_1 I_2 =4 F$
so, the force between the wires will be 4 times the original value.

4 0

3 years ago