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2 years ago

13

A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose an antelope runs 1.5 km with an average speed

of 85 km/h, then runs 0.80 km with an average speed of 67 km/h

1 answer:

Angelina_Jolie [31]2 years ago

4 0

Answer:

10

Explanation:

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In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0

2 years ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

So

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

2 years ago

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

<em> V = v cos Ф </em>

Substituting the values in the above equation

v = 300 x cos 90°

= 300 x 0

= 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

5 0

2 years ago

What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?

erma4kov [3.2K]

<span>E = h x f </span>

<span>. . . then : </span>

<span>f = E / h </span>
<span>f = 4,41•10^-19 / 6,62•10^-34 </span>
<span>f = 6,66•10^14 Hz (s^-1) </span>

<span>b/ What is the wavelength of this light ? </span>
<span>- - - - - - - - - - - - - - - - - - - - - - - - - - - - </span>

<span>λ = c / f </span>
<span>λ = 3•10^8 / 6,66•10^14 </span>
<span>λ = 4,50•10^-7 m </span>

4 0

2 years ago

Read 2 more answers

You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer

boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0

2 years ago