What is a neutral atom?

Look at paths of the projectiles. What effect does air resistance have on the projectile?

inna [77]

Answer:

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction

Explanation:

The projectile motion is described by the kinematics equations giving a parabolic trajectory, where on the x axis there is no acceleration and on the y axis the acceleration is the acceleration of gravity.

When the air resistance is taken into account it can be approximated as a force that opposes the movement that for low speeds is proportional to the speed of the space.

Consequently, the movement in the axis and the acceleration is less, in some cases it can be so small that the constant handle speed, in this case, is called terminal velocity.

On the x-axis the friction force creates an acceleration in the negative direction of the movement that the projectile has to brake.

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction.

6 0

2 years ago

Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat as its temperature starts

slamgirl [31]

Answer:

c = 0.4356 J/gK

Explanation:

Given the following data;

Mass = 450 grams

Initial temperature, T1 = 150°C

Final temperature, T2 = 53°C

Quantity of heat = 34500 Joules

To find the specific heat capacity of the metal;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 53 - 150

dt = -97°C

Converting the temperature in Celsius to Kelvin, we have;

dt = 273 + (-97) = 176 Kelvin

Making c the subject of formula, we have;

$c = \frac {Q}{mdt}$

Substituting into the equation, we have;

$c = \frac {34500}{450*176}$

$c = \frac {34500}{79200}$

c = 0.4356 J/gK

6 0

2 years ago

A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?

kykrilka [37]

We could determine the acceleration using this formula
$\boxed{a= \dfrac{v_{1}-v_{0}}{t} }$

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
$a= \dfrac{v_{1}-v_{0}}{t}$
$a= \dfrac{0-23}{5}$
$a= \dfrac{-23}{5}$
a = -4.6
The acceleration is -4.6 m/s²

8 0

2 years ago

What does displacement describe?

Katarina [22]

<span>A vector which implies that an object has been moved or has changed its position is called displacement. Displacement is usually associated with length and direction of an imaginary straight point. It is the shortest distance from the initial point to the final point of final position (P). Displacement can also be described as the length between the final and initial point on the shortest path. It means an overall change in direction of the object or point of a body.</span>

5 0

3 years ago

Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m

trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ = 30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T: cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction x-axis :

∑Fx = m₁*ax ,ax =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470 -258.77 -300*a =T

T= 1211.23-300*a Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay ,ay =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T: cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) = Equation (2) = T

1211.23-300*a= 980 + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0

3 years ago