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Archy [21]
2 years ago
11

A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?

Physics
1 answer:
Veronika [31]2 years ago
3 0

Answer:

a

Explanation:

its correct

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A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is
Delicious77 [7]

Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
Ever tried to stop a 150-pound (68 kg) cannonball fired towards you at 30 mph (48 km/hr.)? No, probably not. But you may have tr
alex41 [277]
The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
8 0
2 years ago
A baseball player runs 27.4 meters from the
sleet_krkn [62]
6.0 m longer because the player ran 3 and came back 3 at the very end, which looks like he went nowhere but in reality he ran 6.
6 0
2 years ago
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A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver
Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = \frac{distance}{time}

Solution:

The speed of the ball is given by,

speed = \frac{distance}{time}

Thus,

Time = \frac{distance}{speed}

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0
2 years ago
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
11111nata11111 [884]

Answer:

\sigma=0.014\ C/m^2

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2

So, the surface charge density on the sphere is 0.014\ C/m^2.

7 0
2 years ago
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