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1 year ago

2x6573858493.8404? please

Physics

1 answer:

Studentka2010 [4]1 year ago

4 0

Answer:

13,147,716,987.681

if you multiply 2 by 6573858493.8404, you get 13,147,716,987.681

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At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s pe

KonstantinChe [14]

Answer:

4347.8 m/s

Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

$\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB\\$

Take the ratio:

$m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s$

3 0

3 years ago

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

a concrete cube of side 0.50 m and uniform density 2.0 x 103 kg m–3 is lifted 3.0 m vertically by a crane. what is the change in

AlladinOne [14]

Answer:

Change in potential energy = 7350 Joules

Explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube, $d=2\times 10^3\ kg/m^3$

The cube is lifted vertically by a crane to a height of 3 m

We know that, density $d=\dfrac{m}{V}$

So, m = d × V (V = volume of cube = a³)

$m=2\times 10^3\ kg/m^3\times (0.5\ m)^3$

m = 250 kg

We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.

Potential energy at height h is given by :

$PE=mgh$

PE = 250 kg × 9.8 m/s² ×3 m

PE = 7350 Joules

So, change in potential energy of the cube is 7350 Joules.

5 0

3 years ago

Read 2 more answers

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago

In the chemical formula for an ionic compound, which item is written first?

tatuchka [14]

Explanation:

.hahxxjdndjdndjgfndkndidjdodnxondos

5 0

2 years ago