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TEA [102]
2 years ago
5

Two physics students, Ben and Bonnie, are in the weightlifting room. Bonnie lifts the 50 kg barbell over her head (approximately

.60 m) 10 times in one minute; Ben lifts the 50 kg barbell the same distance over his head 10 times in 10 seconds. Which student does the most work? Which student delivers the most power? Explain your answers.
Physics
1 answer:
KatRina [158]2 years ago
5 0

Answer:

Both of them

Explanation:

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On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the
Margaret [11]

Answer:

The change in temperature, \Delta T=9.45^{\circ} C

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from -4^{\circ} F\ to\ 45^{\circ} F in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F

The relation between degrees Celsius and degrees Fahrenheit is given by :

F=1.8C+32

Here, F = 49 degrees

49=1.8C+32\\\\\Delta T=9.45^{\circ} C

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0
3 years ago
Use these options to describe each genotype ​
Sliva [168]
A homozygous dominant
B: homozygous recessive
C: heterozygous
I hope this helps you :)
4 0
3 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of
neonofarm [45]
I hope this would help you

3 0
3 years ago
The y component of the electric field of an electromagnetic wave travelling in the +x direction through vacuum obeys the equatio
notsponge [240]

Answer:

λ =8.57 μ m

Explanation:

Given that

Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C

Standard form  

Ey=Eo cos[k x-ωt]  N/C

By comparing the given equation with the standard wave equation

Eo = 375 N/C

ω  = 2.20 × 10¹⁴ rad/s

We know that ω = 2 π f

f=\dfrac{\omega}{2\pi }

f=\dfrac{2.2\times 10^{14}}{2\pi }\ Hz

f=3.50×10¹³ Hz

We know that the velocity given as

V = f λ

λ =Wavelength

V=Speed = 3 x 10⁸ m/s

\lambda =\dfrac{V}{f }

\lambda =\dfrac{3\times 10^8}{3.5\times 10^{13}}\ m

λ =0.00000857 m              ( 1 μ m = 10⁶ m)

λ =8.57 μ m

8 0
3 years ago
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