All categories

2 years ago

A man walked 50 m to the south, turned, walked 40 m to the west.

Physics

1 answer:

MaRussiya [10]2 years ago

3 0

Answer:

Direction is N 51.3° E

Explanation:

» From trigonometric ratios:

${ \tt{ \red{ \tan( \theta) = \frac{opposite \: { \blue{(south)}}}{adjacent \: { \blue{(west)}}} }}} \\ \\ { \tt{ \tan( \theta) = \frac{50}{40} = \frac{5}{4} }} \\ \\ { \tt{ \theta = { \tan }^{ - 1} ( \frac{5}{4}) }} \\ \\ { \tt{ \theta = 51.3 \degree}} \\ \\ { \boxed{ \tt{ \: direction : \:{ \green{N }}\: 51.3 \degree \: { \green{ E }}}}}$

You might be interested in

Sheila uses a 45N force on her bowling ball across a 15m lane. What work did she do on the bowling ball? Show your work.

Maurinko [17]

Answer:

675J

Explanation:

Given parameters:

Force = 45N

Distance = 15m

Unknown:

Work done by Sheila = ?

Solution:

Work done by a body is the amount of force applied to make a body move through a distance;

Work done = Force x distance

Now;

Work done = 45 x 15 = 675J

6 0

3 years ago

Glaciers begin with snowfall building up and __________________ the ice. (Choose the best answer)

Allisa [31]

Answer:

compacting

Explanation:

i don't think there is very much explanation, the snow falls and compacts the ice to become giant lol

4 0

3 years ago

An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia

Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

$e(t) = e* sin(wt)$

$e(t) = 6.04 * sin(Φ + π)$

$e(t) = 6.04 * sin(32.5 + 180)$

$e(t) = -3.245 V$

Now, Using Kirchoff Voltage Law,

$e(t) - VR- VL - VC = 0$

$-3.24 - 0 - VL - 5.32 = 0$

Finally we have the potential difference across the inductor.

$VL = - 8.56 v$

5 0

3 years ago

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil

Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0

4 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

3 years ago