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Masteriza [31]
3 years ago
13

Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

ada’s velocity to be 8 m/s ˆ I + 3 m/s ˆ j.
What is Jada’s speed with respect to the water?
a. 13.3 m/s
b. 11.3 m/s
c. 8.54 m/s
d. 8.25 m/s
e. 4.24 m/s
f. None of these
Physics
1 answer:
olchik [2.2K]3 years ago
5 0

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

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If the mass of the products measured 120 g, what would be the mass of the reactants? 30 g 60 g 120 g 240 g
Tanya [424]

Answer:

The correct answer option is C

Explanation:

In a balanced chemical reaction mass of the reactant are always equal to mass of the products. Also known as Law of Conservation of Mass  which states that " mass can nor be created nor be destroyed in a chemical reaction."

So, the mass of the reactant will be equal to the mass of products.That is 120 grams.

Hence, the correct answer option(C).

3 0
3 years ago
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2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
3 years ago
Suppose that the acceleration of a model rocket is proportional to the difference between 160 ft/sec and the rocket's velocity.
Levart [38]

Answer:

Explanation:

Given ,

dv / dt = k ( 160 - v )

dv / ( 160 - v ) = kdt

ln ( 160 - v ) = kt + c , where c is a constant

when t = 0 , v = 0

Putting the values , we have

c = ln 160

ln ( 160 - v ) = kt + ln 160

ln ( 160 - v / 160 ) = kt

(160 - v ) / 160 = e^{kt}

1 - v / 160 = e^{kt }

v / 160 = 1 - e^{kt }

v = 160 ( 1 - e^{kt } )

differentiating ,

dv / dt = - 160k e^{kt }

acceleration a   = - 160k e^{kt }

given when t = 0 , a = 280

280 = - 160 k

k = - 175

a = - 160 x - 175 e^{kt}

a = 28000 e^{kt}

when a = 128  t = ?

128 = 28000 e^{kt}

e^{kt } = .00457

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Answer:

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Explanation:

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