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lyudmila [28]
3 years ago
6

A projectile is fired with a velocity of 45 m/s at an angle of 32°. What is the

Physics
1 answer:
Dennis_Churaev [7]3 years ago
8 0

Answer:

C

Explanation:

To calculate adjacent of triangle use

\cos(32)  =  \frac{a}{45}

where 45 is the hypotenuse, and a is the adjacent side (horizontal component)

45 \cos(32)  = 38.16

then round to 38.2

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force
LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

F_y = 51.66 N

F_x = 24.09 N

The net force will be;

F_net = √((F_y)² + (F_x)²)

F_net = √(51.66² + 24.09²)

F_net = √3249.0837

F_net = 57 N

4 0
3 years ago
Awdfwkjbfgkjsenfkjnsekjfgnesklnslkenges
forsale [732]

Answer:

yes

Explanation:

3 0
3 years ago
Read 2 more answers
What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your peri
Papessa [141]
The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!
4 0
3 years ago
A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the
guapka [62]

Answer:

K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

3 0
3 years ago
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