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3 years ago

Do different fruits and vegetables sink or swim

Physics

2 answers:

Artyom0805 [142]3 years ago

4 0

Some fruits would able to float in water.and others can't...because of their mass by each type.

Anni [7]3 years ago

3 0

Some fruits and vegetables sink in water and some float in water. None swim.

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In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

2 years ago

Two resistors R1 = 3 Ω and R2 = 6 Ω are connected in parallel. What is the net resistance in the circuit?

gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

R2 = 6Ω

The net resistance will be:

⇒ $\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}$

On substituting the values, we get

⇒ $\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}$

On taking L.C.M, we get

⇒ $\frac{1}{R_{net}} =\frac{2+1}{6}$

⇒ $\frac{1}{R_{net}} =\frac{3}{6}$

⇒ $\frac{1}{R_{net}} =\frac{1}{2}$

On applying cross-multiplication, we get

⇒ $R_{net}=2 \Omega$

3 0

2 years ago

A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

8 0

2 years ago

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

$\frac{1}{2}mv^{2}=mgh$

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

I hope it helps you!

8 0

3 years ago

HELP ASP PLEASED AND THANKS NO LINKS PLEASED AND NO FILES

SOVA2 [1]

Answer:

just search up a ven-diagram and then try to draw it or trace it then use it for ur question

Explanation:

6 0

3 years ago