All categories

3 years ago

An object is 45 m above the ground when it is dropped. How fast is the object going just before it hits the ground?

1 answer:

3 0

This is a kinematics question.
$v^{2} = v_{0} ^{2} + 2g(y - y_{0}) \\ 
v^{2} = 0 + 2(-9.8)(0 - 45) \\ 
v^{2} = 882 \\ v = \sqrt{882} = 29.7 m/s$

You might be interested in

An 80-kilogram skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles.

ivann1987 [24]

Answer:

The force F is created by the reaction of the Earth to the thrust of the rods, whereby the thrust is created by a force of action and reaction.

Explanation:

To answer this question, let's write Newton's second law of the two axes

Y Axis

Fy + N - W = 0

Fy + N = W

X axis

Fx - fr = 0

Fx = fr

The force F is created by the reaction of the Earth to the thrust of the rods, whereby the thrust is created by a force of action and reaction.

The direction of this force is along the length of the rods that are in an Angle, where the x and y components of the force come from

In general this force is small because the rubbing of the skis is small

8 0

2 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

2 years ago

What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes

mr Goodwill [35]

The current is defined as the amount of charge Q that passes through a given point of a wire in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since I=500 A and the time interval is
$\Delta t=4.0 min=240 s$
the charge is
$Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C$

One electron has a charge of $q=1.6 \cdot 10^{-19}C$ , therefore the number of electrons that pass a point in the wire during 4 minutes is
$N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}$ electrons

3 0

3 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

4 0

3 years ago

The humber bridge in england has the world's longest single span, 1410 m . calculate the change in length of the steel deck of t

Pepsi [2]

Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length

Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m

3 0

2 years ago