Which rocky planet has the strongest magnetosphere?

What is the magnification formula?

erica [24]

Answer:

the height of the image ÷ by the height of the object.

Explanation:

7 0

4 years ago

Two fish swimming in a river have the following equations of motion:

IRINA_888 [86]

Answer:

The second fish, X2, is moving faster than the first fish, X1

Explanation:

The given parameters for the equation of motion of the fishes are;

X1 = -6.4 m + (-1.2 m/s)×t

X2 = 1.3 m + (-2.7 m/s)×t

The given equation are straight line equations in the slope and intercept form, where the slope is the speed and in m/s and the intercept is the starting point of swimming of the fishes

For the first fish, the intercept = -6.4 m, the slope = the speed = -1.2 m/s

For the second fish, the intercept = 1.3 m, the slope = the speed = -2.7 m/s

Whereby the fishes are swimming in the opposite direction of the measurement of length, we have;

The magnitude of the speed of the second fish $\left | -2.7 \ m/s \right | = 2.7 \ m/s$ , is larger than the magnitude of the speed of the first fish $\left | -1.2 \ m/s \right | = 1.2 \ m/s$

Therefore, the second fish, X2, is moving faster than the first fish, X1.

8 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

11. Will the cube in#10 float in water? Will it float in benzene?

densk [106]

Where is the cube I don't see any picture?

5 0

4 years ago

Freight trains can produce only relatively small acceleration and decelerations. (a) what is the final velocity (in m/s) of a fr

m_a_m_a [10]

(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s
Acceleration a = 0.0600 m/s²
time = 8 min = 8 x 60 = 480 s
Final velocity
= u + at
= 5.00 + 0.0600(480)
= 33.8 m/s

The final velocity is 33.8 m/s

5 0

3 years ago

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