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victus00 [196]
3 years ago
12

What is buoyant force?

Physics
1 answer:
Nostrana [21]3 years ago
3 0

This is the upthrust on an object which is placed inside a fluid

This force act upwards and always push upwards

so the correct answer is given as

D. A force within a fluid that pushes upward

this force is always due to pressure difference at two levels of

at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy

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A substance that does not react with metal, has a bitter taste, and turns litmus paper blue is a(n) ___.
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The answer to the question is b because acid is a substance that tastes sour, reacts with metal and turns litmus paper blue
5 0
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Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0
3 years ago
A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the
gregori [183]
The radius, r, of the child from the center of the wheel is
r = 1.3 m

The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
   = (1.3 m)*(1.496 rad/s)
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6 0
3 years ago
Read 2 more answers
Can someone plz answere my question<br> for science
klio [65]
Whats the question?
djdkkd
5 0
2 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
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