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2 years ago

Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

Physics

2 answers:

Aliun [14]2 years ago

7 0

Hope it cleared your doubt.

xenn [34]2 years ago

3 0

We know, a = v₂ - v₁ / t
Here, (v₂-v₁) = 33 - 11 = 22 m/s
t = 10 s

Substitute their values,
a = 22/10
a = 2.2 m/s²

In short, Your Answer would be 2.2 m/s²

Hope this helps!

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While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

Ede4ka [16]

Answer: There are many possible elements, and they are all in the same vertical column as bromine.

Explanation:

In a periodic table, the elements are arranged according to the atomic number. The elements arranged in the same vertical column (known as groups) have same valence configuration and therefore have same chemical properties. Hence, there would be more possible elements having same chemical properties in the same vertical column (group) as Bromine.

7 0

2 years ago

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A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

tankabanditka [31]

Answer

given,

$F_L= 1.52\ N$

$\theta_L= 31^0$

mass of book = 0.305 Kg

so, from the diagram attached below

$F_L cos {\theta_L} + F_b sin {\theta_b} = m g$

$1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8$

$F_b sin {\theta_b} = 2.989 -1.303$

$F_b sin {\theta_b} = 1.686$

computing horizontal component

$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

5 0

2 years ago

Seema knows the mass of basketball. What other information is needed to find the balls potential energy

Lelu [443]

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy $U$ will be:

$U=mgh$

Where:

$m$ is the mass of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity (assuming the ball is on the Earth surface)

$h$ is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

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2 years ago

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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]

<u>Answer:</u>

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

<u>Explanation:</u>

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

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2 years ago

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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago