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3 years ago

Show the correctness of v=fx

Physics

1 answer:

skad [1K]3 years ago

7 0

The answer for the question I believe isF=v/x

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A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

dezoksy [38]

Answer:

For cast iron we have

$h = 0.92 m$

For copper

$h = 1.05 m$

For Lead

$h = 1.23 m$

For Zinc

$h = 2.43 m$

Explanation:

As we know that final speed of the block is calculated by work energy theorem

$W_f + W_g = \frac{1}{2}mv^2$

now we have

$-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2$

now we have

$v^2 = 2gh - 2\mu_k g h cot\theta$

$v = \sqrt{2gh(1 - \mu_k cot\theta)}$

For cast iron we have

$4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}$

$h = 0.92 m$

For copper

$4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}$

$h = 1.05 m$

For Lead

$4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}$

$h = 1.23 m$

For Zinc

$4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}$

$h = 2.43 m$

4 0

3 years ago

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall buildin

OlgaM077 [116]

Solution :

The motion in the y direction.

The time taken by the toy rocket to hit the ground,

$S=ut+\frac{1}{2}at^2$

S = distance travelled = 30 m

u = 0 m/s

a = $9.8 \ m/sec^2$

t= time in seconds

Therefore, $30 =\frac{1}{2}9.8 t^2$

t = 2.47 sec

Now motion in the x direction,

u = 12 m/sec

$a=1.6 t \ m/sec^3$

Upon integration 'v' with respect to 't'

$v=\frac{1.6t^2}{2}+12$

Once again integrating with respect to t,

$s=\frac{1.6t^3}{6}+12 t$

$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$

= 0.0176+29.64

= 29.65 m

Therefore, the toy rocket will hit the ground at 29.65 m from the building.

7 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

kirill115 [55]

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

7 0

3 years ago

Read 2 more answers

An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert o

amid [387]

Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

$F=k*\frac{q1*q2}{r^{2} }$

where:

F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract.
k = is the Coulomb constant or electrical constant of proportionality.
q = value of the electric charges measured in Coulomb (C).
r = distance that separates the charges and that is measured in meters (m).

In this case:

k= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$
q1= 1.602*10⁻¹⁹ C
q2= 1.602*10⁻¹⁹ C
r= 1.2*10⁻⁹ m

Replacing:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }$

and solving you get:

F=1.6*10⁻¹⁰ N

<u><em>The force they will exert on each other is 1.6*10⁻¹⁰ N</em></u>

3 0

3 years ago

Show the correctness of v=fx​

Show the correctness of v=fx