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Eddi Din [679]
3 years ago
9

Which wind blows 30 latitude in both hemisphere almost to the equator

Physics
1 answer:
andriy [413]3 years ago
7 0
Trade winds
Hope it help
You might be interested in
Answer the question in French:
frez [133]

Bonjour

C'est la fille de ma mère mais ce n'est pas moi, qui est-ce ?

<em>She is my mother daughter but, she's not me ? Who is she ?</em>

<em />

<em>Ma sœur  </em><em> = my sister</em>

<em />

<em>☺☺☺</em>

6 0
3 years ago
A person jumps from the roof of a house 3.1-m high. When he strikes the ground below, he bends his knees so that his torso decel
Sveta_85 [38]
Part A

Free fall motion
h = 3.1 m

Equation: Vf = √(2gh) = √(2*9.8 m/s^2 * 3.1 m) = 7.8 m/s

That is the only part in the question.
3 0
3 years ago
Hey stob it.<br> Please help me.<br> Cmon help me.<br> Plz.
Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0
3 years ago
after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF
Snezhnost [94]

Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

4 0
3 years ago
A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

F_s = \mu N

where \mu is the coefficient of friction and N is the normal force.

When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

N  = Mg- F_psin(\theta)

and hence the frictional force becomes

F_s = \mu (Mg- F_psin(\theta)).

Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em>W=mg<em> is independent of the external forces on the block. </em>

6 0
3 years ago
Read 2 more answers
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