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Archy [21]
3 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]3 years ago
3 0

Answer:

a

Explanation:

its correct

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Answer:

MRBEAST-

Explanation:

6 0
3 years ago
Read 2 more answers
A bird with a mass of 0.88 kg has a potential energy of 96 J. How high off the ground is the bird?
Ray Of Light [21]

Answer:

Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?

Explanation:

solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).

5 0
3 years ago
The electromagnetic waves that have the lowest frequencies are called
o-na [289]
A. Radio waves
Have the lowest frequencies
4 0
3 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
Art [367]

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

8 0
3 years ago
If the radius of a blood vessel drops to 84.0% of its original radius because of the buildup of plaque, and the body responds by
erma4kov [3.2K]

To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.

From Bernoulli it is possible to express the change in pressure as

\Delta P = \frac{1}{2}\rho (v_1^2-v_2^2)+ \rho g (h_1h_2)

Where,

v_i =Velocity

\rho = Density

g = Gravitational acceleration

h = Height

From the given values the change of flow is given as

R = r^4P

Therefore between the two states we have to

\frac{R_2}{R_1} = \frac{r_2^4 P_2}{r_1^4 P_1} *100\%

\frac{R_2}{R_1} = \frac{84^4 (110)}{100^4*(100)} *100\%

\frac{R_2}{R_1} = 54.77\%

The flow rate will have changed to 54.77 % of its original value.

4 0
2 years ago
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