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2 years ago

6

What is common to the terrestrial planets? A. equal rotational periods around the Sun B. thin crust and a dense iron core C. sol

id outer layer and an atmosphere D. a solid outer layer composed of rock and minerals

1 answer:

Darya [45]2 years ago

7 0

Answer:

I would say D. All terrestrial planets are made up of rock and minerals and the other planets are made up of that do not have a solid surface.

Answer D.

Explanation:

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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

Nat2105 [25]

Answer:

Explanation:

Given

diameter $d=20 \mu m$

density $\rho =1300 kg/m^3$

frequency $\nu =150 Hz$

Length of silk strand $L=14 cm$

Velocity in the string is as follows

$\nu =\sqrt{\frac{T}{\mu }}$

The expression for Fundamental Frequency

$f=\frac{\nu }{2l}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}$

$f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}$

Squaring

$f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}$

$T=4\rho \cdot A\cdot l^2\cdot f^2$

$T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2$

$T=7.2\times 10^{-4} N$

8 0

3 years ago

Suppose that the magnetic field in some region has the form B = kz ˆx (where k is a constant). Find the force on a square loop (

AleksAgata [21]

Answer:

$F = ika^2$

Explanation:

As we know that loop is placed in YZ plane and magnetic field is along x direction

So here net force on the side of the loop which lies along Y axis is given as

$F_1 = i (\vec L \times \vec B)$

here we know that on Y axis z = 0

so B = 0

so we have

$F_1 = 0$

now on the opposite side we have z = a

so magnetic field is given as

$B = ka$

so force on that side is given as

$F = i(\vec L \times \vec B)$

$F = i(a)(ka) sin90$

$F_2 = ika^2$

so net force on the loop is given as

$F = F_1 + F_2$

$F = ika^2$

5 0

3 years ago

The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic

Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight = $\frac{2uSin(A)}{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = $\frac{2(4)Sin(60)}{10}$ = 0.7

ii) distance travel = Range = R = $\frac{u^{2}Sin(2A) }{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = $\frac{4^{2}Sin(2*60) }{10}$ = 1.39

iii) Maximum Height = H = $\frac{u^{2}(Sin(A))^{2} }{2g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: $\frac{4^{2}(Sin60)^{2} }{2*10}$ = 0.6

4 0

3 years ago

4) A drag racer starts her car from rest and accelerates at 10.0 m/s² for a distance of 400 m (1/4 mile). (a) How long did it ta

mel-nik [20]

Answer:

A) s=1/2at^2

t=√(2s/a)=√(2x400)/10.0)=9.0s

B) v=at

v=10.0x9=90m/s

3 0

3 years ago

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

3 years ago