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Stels [109]
3 years ago
6

Using -10 m/s2 for acceleration due to gravity, what would be the total displacement of the object if it took 8 seconds before h

itting the water?
Physics
1 answer:
ASHA 777 [7]3 years ago
6 0
If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres
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If a circuit contains a 12v battery and two 3 bulbs in series
Paha777 [63]

If the resistance of each bulb is 3 ohms, and they are connected across the battery, then current will flow through them.

In series, the two 3-ohm bulbs will look like a single 6-ohm resistance.

      Current = (voltage) / (resistance)  =  (12v) / (6 ohms) = 2 Amperes.

The battery will need to deliver some power to the circuit.

       Power = (voltage) x (current) = (12v) x (2 Amperes)  =  24 watts .

Each bulb will dissipate 12 watts, in the form of light and heat.

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6 0
4 years ago
A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
4 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one
Llana [10]
We don't know how many of ANY color are in the bag right now, so there's no way to calculate an answer.

What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.
4 0
3 years ago
Froghopper insects have a typical mass of around 11.1 mg 11.1 mg and can jump to a height of 55.3 cm. 55.3 cm. The takeoff veloc
Molodets [167]

Answer:

a = 2710m/s²

Explanation:

See attachment below.

4 0
3 years ago
Read 2 more answers
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