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4 years ago

Select the correct answer from each drop-down menu.

Physics

1 answer:

kirill115 [55]4 years ago

8 0

Answer:

Explanation:

I think but I am sure

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A car travels at a constant speed around a circular trackwhose

Mashutka [201]

Answer:

a = 0.77 m/s^2

Explanation:

The centripetal acceleration is given as

$a = \frac{v^2}{R}$

The velocity of the car can be calculated using the circumference of the track and the period of the car.

$2\pi R = v T\\2\pi 2.6 = v 360\\v = 0.045 ~km/s = 45~m/s$

So, the acceleration is

$a = \frac{45^2}{2.6\times 10^3} = 0.77~m/s^2$

8 0

4 years ago

Which branch of science deals with the study of the structures shown here?

Arturiano [62]

Vertebrate Zoology, I’m pretty sure

3 0

3 years ago

Read 2 more answers

Mya (m = 65 kg) floats 4 m from her spaceship (m = 1600 kg). What gravitational force would each mass exert on the other?

Aleksandr-060686 [28]

Answer:

ans: 4.34 × 10^(-9) N

Explanation:

mass of Mya say (m) = 65 kg

mass of spaceship say (M) = 1600 kg

universal gravitational constant(G) =6.67 × 10^(-11) Nm²/kg²

separation distance (d) = 4m

so,

gravitational force (F)= GMm/d²

=( 6.67 × 65 × 1600) / ( 10¹¹ × 4²)

= 4.34 × 10⁴ / 10¹³

= 4.34 × 10^(-9) N

8 0

3 years ago

A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j

fenix001 [56]

Answer:

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

$\vec F_e = q\vec E$

$\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3$

$\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j$

Now force on moving charge due to magnetic field is given as

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

$\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )$

$\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i$

$\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k$

Now net force due to both

$F = F_e + F_b$

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

7 0

3 years ago

1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the minimum coefficient of static f

tatuchka [14]

Answer:

C. 0.1

Explanation:

5 0

3 years ago