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3 years ago

Active immunity is passed on from a mother to her baby.

1 answer:

3 0

True.The immune system in babies. Antibodies are passed from mother to baby through the placenta during the last three months of pregnancy.

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What happens to the magnitude of the momentum of an object if the mass of the object halved?<br>

nekit [7.7K]

Answer:

Momentum is directly proportional to velocity. If momentum of an object is doubled , but its mass does not increase (so velocity remains below the speed of light) then its velocity is doubled. If the velocity is doubled, then the kinetic energy increases by four times.

Explanation:

Hope it is helpful....

8 0

3 years ago

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial:

$0.8\cdot t^{2}-56.92\cdot t +3.872 = 0$ (3)

$t_{1} \approx 71.082\,s$ , $t_{2}\approx 0.068$

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

3 0

3 years ago

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine o

noname [10]

Answer:

The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Explanation:

Using the equations of motion,

When the rocket is fired from the ground,

u = initial velocity = 0 m/s (since it was initially at rest)

a = 10 m/s²

The engine cuts off at y = 0.5 km = 500 m

The velocity at that point = v

v² = u² + 2ay

v² = 0² + 2(10)(500) = 10000

v = 100 m/s

The velocity at this point is the initial velocity for the next phase of the motion

u = 100 m/s

v = final velocity = 0 m/s (at maximum height, velocity = 0)

y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?

g = acceleration due to gravity = - 9.8 m/s²

v² = u² + 2gy

0 = 100² + 2(-9.8)(y)

- 19.6 y = - 10000

y = 510.2 m = 0.510 km

So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Hope this helps!!!

5 0

3 years ago

1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line

poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

$v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s$

Hence, his average velocity is 10 m/s.

6 0

3 years ago

A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc

jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

$V_{ox}=V_o\cos \theta$

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

$\begin{gathered} V_{ox}=2.70\cos 0 \\ \\ V_{ox}=2.7\text{ m/s} \end{gathered}$

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

$V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt$

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

$\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\ \\ V_{oy}=2.70\sin 0-9.8(2.10) \\ \\ V_{oy}=0-20.58 \\ \\ V_{oy}=-20.58\text{ m/s} \end{gathered}$

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

6 0

1 year ago