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3 years ago

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An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

15 minutes = 0.25 hours)

1 answer:

Simora [160]3 years ago

4 0

..... It would possibly she eenejjsjejeej 1.4

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A car starts from rest and after 20 seconds it's velocity becomes 108km find the acceleration of the car

andre [41]

Answer:

1.5 km/s²

Explanation:

Given that:

a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s

after time (t) = 20 seconds

the final velocity = 108 km/hr = 30 m/s

The acceleration (a) of the car can be determined by using the formula:

$a = \dfrac{v-u}{t}$

$a = \dfrac{30\ m/s -0 \ m/s}{20 \ s}$

$a = \dfrac{30 \ m/s}{20 \ s}$

a = 1.5 km/s²

7 0

3 years ago

A 32-kg child decides to make a raft out of empty 1.0-L soda bottles and duct tape. Neglecting the mass of the duct tape and pla

Vedmedyk [2.9K]

Answer:

32 bottles

Explanation:

If we create a free body diagram on the child we have his weight and the bouyant force

W-B=0

They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced

W=B=Ww

Mg=mg

32 kg=mass of water displaced

1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles

6 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

If a person is 6.25 m away from a 60.0 w speaker, what is the sound level they are hearing?

vodka [1.7K]

Answer:

110.87 dB

Explanation:

(I got it right on Acellus)

I= P/4(pi)r^2 = 60/4(pi)6.25^2

60/4(pi)6.25^2=0.12223

B=10log(I/Io)

B=10log(0.12223/1*10^-12) = 110.87 dB

111 in sigfigs

4 0

2 years ago

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