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horsena [70]
3 years ago
9

HELP with this please.

Physics
2 answers:
jasenka [17]3 years ago
6 0
17. C. Isotopes
18. Tracers
19. Alpha Decay
jolli1 [7]3 years ago
4 0
17. C. 
18. Tracers
19. Alpha Decay

Hope this helps and please mark as brainliest!
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A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.
leva [86]
This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)
6 0
4 years ago
Calculate Vector component in Y if the hypotenuse is 32 and angle is 45
Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

v_y = v \times sin(\theta)

Substituting the values from the question gives;

v_y = 32 \times sin(45^{\circ}) \approx 22.6

The vector component in the y-axis, v_y, is approximately 22.6.

8 0
3 years ago
n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of
lilavasa [31]

Answer:

  B : is independent of the natural frequency of the oscillator

Explanation:

You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.

The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.

Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.

5 0
3 years ago
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim
Nonamiya [84]

Answer:

The distance covered by puck A before collision is  z = 8.56 \ m

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  v_A =  3.90 \ m/s

        The speed of puck B is  v_B  =  4.30 \ m/s

The distance covered by puck A is mathematically represented as

     z =  v_A * t

  =>  t  =  \frac{z}{v_A}

 The distance covered by puck B  is  mathematically represented as

      18 - z =  v_B  * t

=>   t  = \frac{18 - z}{v_B}

Since the time take before collision is the same

        \frac{18 - z}{V_B}  =  \frac{z}{v_A}

substituting values

          \frac{18 -z }{4.3}  = \frac{z}{3.90}

=>      70.2 - 3.90 z   = 4.3 z

=>       z = 8.56 \ m

8 0
3 years ago
The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
Gemiola [76]

Answer:

6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

I=0.040 \mu A=4\cdot 10^{-8}A

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T

5 0
3 years ago
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