Ways in which any potential risk to the patient can be minimized include

MacGyver is riding on a skateboard and moving west of the velocity of 7 m/s. Macgyver has a mass of 92 kg and the skateboard has

ivann1987 [24]

Answer:

64.5

Explanation:

(m1 + m2) v initial = (m1 * v1 final) + (m2 * v2 final)

(92 + 8)(-7) = (92 x 2) + (8 x v2 final)

(100)(-7) = (184) + (8 x v2 final)

-700 = (184) + (8 x v2 final)

-516 = (8 x v2 final)

-516/8 = v2 final

-64.5 = v2 final

8 0

3 years ago

A(n) ______ is matter that is composed of one type of atom. answerd::: An Element

OverLord2011 [107]

The answer would me an element

7 0

4 years ago

Read 2 more answers

Transcranial magnetic stimulation (TMS) is a noninvasive method for studying brain function, and possibly for treatment as well.

lubasha [3.4K]

Answer:

525 V

Explanation:

A = Area = $1.75\times 10^{-2}\ m^2$

$\dfrac{dB}{dt}$ = Rate of change of magnetic field = $3\times 10^4\ T/s$ (assumed)

Induced electromotive force is given by

$E=A\dfrac{dB}{dt}\\\Rightarrow E=1.75\times 10^{-2}\times 3\times 10^{4}\\\Rightarrow E=525\ V$

The induced electromotive force is 525 V

3 0

3 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0

4 years ago

Read 2 more answers

When the arrow is released, how does the the force on the arrow (by the bow) compare to the force on the archer (by the bow)?

Jet001 [13]

Answer:

hello some parts of your question is missing below is the complete question

A 60.0 kg archer standing on a friction-less ice shoots a 100g arrow at a speed of 80.0 m/s to the right .When the arrow is released, how does the the force on the arrow (by the bow) compare to the force on the archer (by the bow)?

Answer : The force on the arrow is larger than the force on the archer because the arrow has a higher final velocity

Explanation:

we can compare The force on the arrow to the force on the archer by using the final velocities of the archer and the arrow ,hence The force on the arrow is larger than the force on the archer because the arrow has a higher final velocity and this is according to Newton's law which states for a every action there is an equal reaction i.e the force applied on the arrow will be the opposite of the force on the archer

5 0

3 years ago