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lubasha [3.4K]
3 years ago
15

What is the approximate velocity of the object at 5 seconds ? .

Physics
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

do you have an image?

Explanation:

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Caleb is filling up water balloons for the Physics Olympics balloon tosscompetition. Caleb sets a 0.50-kg spherical water balloo
Mashcka [7]

a)

• P = F/A

P = pressure = 630 N/m^2

F = force

A = area

F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N

m= mass

g= gravity

P = F/A

A = F/P

A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2

b)

• Area of a circle = pi* radius ^2

7.778 x 10^-3 m^2 = pi* radius ^2

√(7.778 x 10^-3 m^2 / pi ) = radius

radius = 0.04976 m

Answers:

a ) 7.778 x 10^-3 m^2

b) 0.04976 m

8 0
1 year ago
Why isn't pluto considered a planet anymore?
aivan3 [116]

The answer is. It did not meet the three criteria the IAU uses to define a full-sized planet.

1. It is in orbit around the Sun

2. It has sufficient mass to assume hydrostatic equilibrium

3. It has 'cleared the neighborhood' around its orbit

Pluto has not 'cleared its neighborhood'

3 0
3 years ago
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?
Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u =  6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0
3 years ago
A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that
FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

t=\sqrt{\dfrac{2s}{g}}

t=\sqrt{\dfrac{2\times 32}{9.8}}

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

v=\dfrac{d}{t}

v=\dfrac{11\ m}{2.55\ s}

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0
3 years ago
I'm sorry if any one seen that post about me saying I'm done with life and I don't want to be here it was my little cuzin @vicky
solniwko [45]
I did not see the post. But if she actually needs help She can dm me. keep y'all's head up❤
8 0
3 years ago
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