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3 years ago

VI. A student attaches four drawing pins to a copper rod using candle wax as shown in the diagram

Physics

2 answers:

madreJ [45]3 years ago

8 0

Answer:

mzlapqnxkswoncksowncow

mestny [16]3 years ago

5 0

Explanation:

$C.) \: Conduction \: \\ \\ = > it \: is \: a \: process \: of \: transferring \: \\ \\ of \: heat \: from \: one \: object \: to \: another \\ \\ when \: they \: are \: in \: contact \:$

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From

madreJ [45]

Answer:

From C to D, kinetic energy is transformed into

gravitational potential energy.

4 0

3 years ago

A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a

worty [1.4K]

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

7 0

3 years ago

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

8 0

3 years ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Svetlanka [38]

Answer:

Mass of base ball $m_b=3.992*10^{14}kg$

Explanation:

Circumference of baseball = 2πr = 23 cm

So radius of baseball = 3.66 cm = $3.66*10^{-2}$ m

Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

Mass of proton = $10^{-27}$ kg

Diameter of proton = $10^{-15}$ m

Radius of proton = $5*10^{-16}$ m

Volume of ball = $\frac{4}{3} \pi r^3$

Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

So mass of base ball $m_b=3.992*10^{14}kg$

5 0

4 years ago

Read 2 more answers

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

Σ $F_{y}$ =0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

$N-W=0$

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

$f_{k}=N$ μ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

$f_{k}=2450N*0.12$

$f_{k}=294$

With this information we can go ahead and find the sum of horizontal forces:

Σ $F_{x}$ =ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N- $f_{k}$ =ma

750N-294N=(250kg)a

when solving for a we get:

$a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}$

so the crate's acceleration is 1.82m/s².

5 0

3 years ago