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3 years ago

Which of the following are examples of projectile not

Physics

1 answer:

fgiga [73]3 years ago

4 0

Answer:

Examples

Aeroplane taking a flight.
Triggering a bullet in sky.
Kicking a ball in sky.

Non-Examples

A moving vehicle.
A running ball.

Explanation:

Projectile motion is the motion of a body which moves due to its inertia and the gravitational force since it takes place in a gravitational field.

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3 years ago

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On Earth, all objects dropped in freefall will:

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3 years ago

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement fr

Lady_Fox [76]

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

<h3>What is mechanical energy?</h3>

Mechanical energy is the sum of potential energy and kinetic energy.

Total mechanical energy = P.E max = K.E max

Total mechanical energy = K.E +P.E

Given is the kinetic energy is (3/4)E.

E= (3/4)E + P.E

P.E = (1/4) E

Maximum potential energy =E = (1/2) k A²

Here. A is the maximum displacement and k is the spring constant.

The potential energy at kinetic energy of (3/4) E is

P.E = (1/4)E = (1/8) k A²

Therefore, the system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

Learn more about mechanical energy.

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6 0

3 years ago

A car has a mass of 1520 kg. While traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.

docker41 [41]

Answer:

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Known data

m=1520 kg : mass of the car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction downward

FN : Normal force : In vertical direction upward

f : Friction force: In horizontal direction

Calculated of the W

W= m*g

W= 1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a = (-5958.4) / ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁ = (20)²

d₁ = (20)² / (7.84)

d₁ = 51.02 m

(b) Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂= (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N

FN₂= (1.5)*14896 N

f= 0.4* (1.5)*14896 N

a = - f/m₂ = - 0.4* (1.5)*14896 N /(1.5) *1520

a = -3.92 m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92 m/s²

d₂ = d₁ = 51.02m

6 0

4 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago