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marusya05 [52]
2 years ago
11

Find the horizontal component and the vertical component ​

Physics
1 answer:
aleksandr82 [10.1K]2 years ago
8 0

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

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A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
Which of the following is the equation for calculating daily caloric needs? A. BMR calories + Digestion calories B. BMR calories
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Explanation:

To determine your total daily calorie needs, multiply your BMR by the appropriate activity factor, as follows: If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2. If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.

<em><u>HAPPY TO HELP</u></em>

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I hope the answer is D. Pressure

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Naddika [18.5K]
Oxygen....................
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3 years ago
The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob
Naddik [55]

(a) 0.2888 kg m^2

The moment of inertia of a uniform-density disk is given by

I=\frac{1}{2}MR^2

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2

(b) 142.5 J

The rotational kinetic energy of the disk is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s

And so, the rotational kinetic energy is

K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J

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The rotational angular momentum of the disk is given by

L=I\omega

where

I is the moment of inertia

\omega is the angular velocity

Substituting the values found in the previous parts of the problem, we find

L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s

8 0
3 years ago
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