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viktelen [127]
3 years ago
15

You plan to take your hair blower to europe, where the electrical outlets put out 240 v instead of the 120 v seen in the united

states. the blower puts out 1600 w at 120 v .
Physics
1 answer:
Dmitrij [34]3 years ago
8 0
P = U × I
I = P / U = 1600W / 120V = 13.4A
P = 240V × 13.4A = 3216W

If your hair blower isnt rated for 220- 230V(this is the voltage in EU) you are most likely going to burn it.
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In the diagnostic radiology energy range (which includes mammography) from 23 to 150 kVp, which of the following tissues possess
Semmy [17]

Answer:

Bone

Explanation:

Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.

The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

4 0
3 years ago
Is it possible to apply the same amount of force and do different amounts of work?
Tamiku [17]
Yes, with simple machines
5 0
3 years ago
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancin
dimulka [17.4K]

Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

Comparing

mi = mv + 1.41 m j

v = i - 1.41 j

Magnitude of velocity

      v=\sqrt{1^2+(-1.41)^2}=1.73m/s        

Direction,  

       \theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0             

Velocity is 1.73 m/s along 54.65° south of east.

5 0
3 years ago
A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w
SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

F=ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

8 0
4 years ago
At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.9 m/s toward the h
forsale [732]

Answer:

17.10 m

Explanation:

When the toboggan stops, we have:

N-w_y=0\\F-w_x=0

The x-component of weight is the product of the weight and the sine of the angle above the horizontal, so the y-component of W is the product of the weight and the cosine of the angle above the horizontal.

N-mgcos(25^\circ)=0\\F-mgsin(25^\circ)=0\\F=mgsin(25^\circ)(1)

The work-energy principle states that the change in the kinetic energy of an object is equal to the net work done on the object.

\Delta K=W\\K_f-K_i=F\cdot h\\\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}=Fhcos(180^\circ)

Recall that v_f=0, replacing (1):

-\frac{m(v_i)^2}{2}=mgsin(25^\circ)h(-1)\\h=\frac{(v_i)^2}{2gsin(25^\circ)}\\h=\frac{(11.9\frac{m}{s})^2}{2(9,8\frac{m}{s^2})sin(25^\circ)}\\h=17.10 m

7 0
3 years ago
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