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skelet666 [1.2K]

3 years ago

8

What atoms of a specific element that have the same number of protons but a different number of neutrons are called:?

2 answers:

Ne4ueva [31]3 years ago

5 0

Atoms of the same element that have a different number of neutrons are called isotopes.isotopes have the same atomic number but different atomic masses.

sergiy2304 [10]3 years ago

4 0

Answer:

isotopes

Explanation:

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atoms are the principle constituent of ______ A. all matter b. only solids liquids and gases c. all subatomic particles d. only

Soloha48 [4]

I think it is a. all matter

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3 years ago

A rock is thrown horizontally from a bridge with a speed of 29.0 m/s. if the rock is 23.7 meters above the river at the moment o

mash [69]

It would be 1.5 meters im sure form that distance to me is that nswe

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3 years ago

A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?

nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0

3 years ago

What force must the worker exert to get the box moving & what force must the worker exert to accelerate the box at 0.1 meter

photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

$F - F_k = m*a$

here

$F_k = kinetic friction = 220 N$

$m = mass = 500 kg$

$a = acceleration = 0.1 m/s^2$

now we will have

$F - 220 = 500* 0.1$

$F = 220 + 50 = 270 N$

3 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

6 0

3 years ago