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Alenkasestr [34]

4 years ago

6

Esmerelda may have a brain tumor. Her doctor will most likely use to view the tumor.

2 answers:

Furkat [3]4 years ago

7 0

A diagnosis. Your doctor will give you a physical exam if you have symptoms of a brain tumor followed by a neurological exam and then an MRI scan

ludmilkaskok [199]4 years ago

7 0

The doctor is likely to use an MRI Scan

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What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters

Setler79 [48]

Answer:A

Explanation:I TOOK THE Test

5 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

4 years ago

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall

kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

$F = q(\vec v \times \vec B)$

so here we have

$F = qvB sin90$

since force is perpendicular to the velocity so here it must be centripetal force

here we have

$\frac{mv^2}{R} = qvB$

so we have

$R = \frac{mv}{qB}$

$R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}$

$R = 4.06 cm$

so here it will move in circle with radius 4.06 cm

8 0

3 years ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0

3 years ago

If the discovery of oxygen wasn’t made in 1774, do you think it would have been discovered later on?

Scilla [17]

I don’t know sorry I’m new

8 0

2 years ago