Does the boat hold more weight in the regular water or the salt mixture

This is my question. I need the answer stat.

Kipish [7]

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

5 0

1 year ago

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What is the Acceleration of a car traveling 25 m/s and in 5 seconds<br> traveling 65 m/s? *

WARRIOR [948]

Answer:

8m/s²

Explanation:

a = change in v/t

a = (65-25)/5

a = 8

7 0

2 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

2 years ago

What do solutions and colloids have in common

Andreyy89

I believe thye answer is either d or c

6 0

2 years ago

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A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

Komok [63]

Answer:

The fastest satellite must change orbit

The most massive body (m₁) transfers more momentum to the satellite,

Explanation:

For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants

Most massive body (m1)

initial. Before the crash

p₀₁ = M v + m₁ v₁

after the crash

$p_{f1}$ = M v´ + m₁ v₁´

how momentum is conserved

p₀ = p_{f}

Lighter body (m2)

p₀₂ = M v + m₂ v₂

p_{f2} = M v´ + m₂ v₂´

Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.

We can see that p₀₁> p₀₂

Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.

The fastest satellite must change orbit, thus rotating at a different distance from Earth

5 0

2 years ago