Please help on this one?

Which type of physical activity is being performed in the picture?

mafiozo [28]

B strength training I think that’s the answer

3 0

3 years ago

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration $\frac{mv^{2}}{r}$

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

6 0

3 years ago

Which of the following means that an image is located behind a lens? A. +di B. -di C. -do D. +do

STatiana [176]

For a lens, the following sign convention is generally used:

- f (the focal length) is positive for a converging lens and negative for a diverging lens
- $d_o$ (the distance of the object from the lens) is positive if the object is in front of the lens and negative if it is behind the lens
- $d_i$ (distance of the image from the lens) is positive if the image is behind the lens (real image) and negative if the image is in front of the lens (virtual image)

Therefore, the correct option is
<span>A. +di
</span>which mens that the image is real and located behind the lens.

4 0

3 years ago

A 2-kg box is pushed to the right by a force of 4 N for a distance of 32 m. It has an initial velocity of 4 m/s to the right. NO

rewona [7]

Answer: a) 8 Kg m/s b) 16 Kg m/s c) 24 Kg m/s d) 16 J e) 128 J f) 144 J

g) 4 s

Explanation:

a) As momentum by definition is the product of mass times the velocity (is a vector quantity), we can write in this case the following:

pi = m. v₀ = 2 Kg . 4 m/s = 8 Kg. m/s

b) In order to get the change in momentum, we need to get first the final speed of the object.

As we have the total distance travelled, and we could find the acceleration, we could use a kinematic equation to solve the question, but later we will need the kinetic energy, it would be better to apply the work-energy theorem, and calculate ΔK as the work done by external force F, as follows:

ΔK = F . d = 1/2 m (vf² - v₀²)

As we know F, d, m, and v₀, we can solve the equation above for vf:

vf = 12 m/s

So, we can compute the final momentum as follows:

pf = m. vf = 2 Kg. 12 m/s = 24 Kg. m/s

Finally, we can find the change in momentum, as the difference between the final momentum and the initial one, calculated in a):

Δp = pf - pi = 24 Kg. m/s - 8 Kg. m/s = 16 Kg. m/s

c) As we have already found, final momentum is as follows:

pf = m . vf = 2 Kg. 12 m/s = 24 Kg. m/s

d) By definition the initial kinetic energy of the box is as follows:

Ki = 1/2 m v₀² = 1/2. 2 Kg .4² m²/s² = 16 J

e) We can find the change in the kinetic energy taking directly the difference between the final and initial ones, as follows:

ΔK = Kf - Ki = 1/2. 2 Kg (12² - 4²) m²/s² = 128 J

f) From above, we have Kf = 1/2 m. vf² = 1/2 . 2 Kg. 12² m²/s² = 144 J

g) As we know the magnitude of F, and the value of m, we can find the acceleration (assumed constant) , applying Newton's Second Law, as follows:

Fext = m .a ⇒ a = F/m = 4 N / 2 Kg = 2 m/s²

Appying the definition of acceleration, we can solve for t, as follows:

t = (vf-v₀) / a = (12 m/s - 4 m/s) / 2 m/s² = 4 s

6 0

3 years ago