With what force will a car hit a tree if the car has a mass of 3,000 kg and it is accelerating at a rate of 2m/s2

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

olchik [2.2K]

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.

5 0

3 years ago

. If force (F), work (W) and velocity (v) are taken as fundamental quantities.

alex41 [277]

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

7 0

3 years ago

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If a car is traveling at a speed of 45 miles per hour, how far can it travel in 40 minutes?

dusya [7]

Answer:

30 miles

Explanation:

Step 1:

Divide -> 45/60= .75 miles/minute

Step 2:

Multiply -> .75 x 40= 30

3 0

3 years ago

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A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

2 years ago

HELP PLEASE!!!!

Archy [21]

Suppose you mixed two chemicals in the lab until you could not tell the two apart. After some time passed, a white powder formed which would not dissolve, and settled on the bottom. The mixture was first homogeneous then heterogeneous.

6 0

3 years ago

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