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2 years ago

1. If I dig a 6FT hole how deep is that hole?

Physics

2 answers:

iren [92.7K]2 years ago

5 0

Answer:

For the first one its about 25 feet

Explanation:

galina1969 [7]2 years ago

5 0

Answer:

1~6ft.

2~ 10

3~ Echo

4~ A photo shoot

5~ talk

6~ nose

7~tent

8~popcorn

Explanation:

</3 PureBeauty

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A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar

tekilochka [14]

<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

$\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m$

Magnitude of his displacement = 2729.47 m

3 0

3 years ago

Read 2 more answers

A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in

seropon [69]

Answer:

Magnetic force, $F = 3.52\times 10^{-13}\ N$

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, $q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C$

Speed of the ion in the magnetic field, $v=5\times 10^6\ m/s$

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

$F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N$

So, the magnitude of magnetic force on the ion is $3.52\times 10^{-13}\ N$ .

3 0

3 years ago

Read 2 more answers

A 20-N force acts on an object with a mass of 2.0kg. what is the objects acceleration?

ale4655 [162]

Answer:

10 m/s^2

Explanation:

Equation: F = ma.

a = acceleration

m = mass

F = force

Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.

F = 20

m = 2

a = ?

a = F/m

a = 20/2

a = 10 m/s^2

7 0

3 years ago

Read 2 more answers

Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto

Zina [86]

Elastic

Explanation is that it is force which is snapping back

7 0

3 years ago

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A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$

Our values are:

$V=30.1m/s$

$C_d=0.45$

$A=3.3m^2$

$\rho=1.2kg/m^3$

Replacing,

$F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2$

$F_D=807.25N$

We need calculate now the energy lost through a time T, then,

$W = F_D d$

But we know that d is equal to

$d=vt$

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

$W=(807.25N)(30.1m/s)(3600s/1hr)$

$W=87.47*10^6J$ (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0

3 years ago