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3 years ago

A speed-time graph is shown below:

Physics

1 answer:

VladimirAG [237]3 years ago

5 0

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

$a_{ave}=0.5 m/s^{2}$

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What is the best use of an atomic model to explain the charge of the particles in thomson's beams

hoa [83]

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation:

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

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3 years ago

What type of energy does friction produce?

VMariaS [17]

Answer:

The correct option is A = adiant negy.

6 0

3 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the

Tems11 [23]

Mass=m=20kg
Radius=r=4m
Velocity=10m/s=v

We know

$\boxed{\sf F_c=\dfrac{mv^2}{r}}$

$\\ \sf\longmapsto F_c=\dfrac{20(10)^2}{4}$

$\\ \sf\longmapsto F_c=\dfrac{20(100)}{4}$

$\\ \sf\longmapsto F_c=\dfrac{2000}{4}$

$\\ \sf\longmapsto F_c=500N$

6 0

3 years ago

A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0

4 years ago

Please help!!

Ronch [10]

Answer:

Explanation:

The distance will be the total distannce covered during the journey.

If you move 3 meters East and move 4 meters north, then the distance will be calculated as;

Distance = distnace through East+distance through north

Distance = 3m + 4m

Distance = 7m

Displacement is the distance covered in a specified direction. It is the shortest distance covered by me. This can be gotten using the Pythagoras theorem.

d² = 3²+4²

d² = 9+16

d² = 25

d = √25

d = 5m

Hence the displacement of the object is 5metres

5 0

3 years ago