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3 years ago

A student is given a sample of matter and asked to determine whether it is an element.

Physics

2 answers:

Pepsi [2]3 years ago

5 0

I think the correct answer is C

scoundrel [369]3 years ago

4 0

It's a mixture cause the 3 different pure substances can only be separated if its a mixture

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A biker goes up hill with a constant speed of 10km/h. Then he goes downhill with a constant speed of 50km/h. What's his average

Sladkaya [172]

This is not as simple as it looks.

His average speed is NOT (10km/hr + 50km/hr)/2 = 30 km/hr.

You have to use the definition of speed:

Speed = (total distance covered) / (time to cover the distance).

Let's say the distance up (and down) the hill is 'd' .

Then the time it takes to go up the hill is (d/10) hours.

And the time it takes to come down the hill is (d/50) hours.

Total distance = 2d km

Total time = (d/10) + (d/50) = (5d/50) + (d/50) = 6d/50

Speed = distance/time = 2d/(6d/50) = 100d/6d

Speed = 100/6 = 16-2/3 km/hr

4 0

3 years ago

Define projectile in your own .

lorasvet [3.4K]

Answer:

a body which was thrown in space ,moves under the influence of gravity only is defined as projectile.

5 0

3 years ago

Read 2 more answers

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

Which object would have a LARGER gravitational force acting upon it? (assume the objects are at the same height above the Earth.

djverab [1.8K]

A) 5 kg block of wood
B) 100 kg person
C) 412 kg motorcycle
D) 2540 kg elephant
Answer: D

7 0

3 years ago

Read 2 more answers

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$