Can someone please help?

andrezito [222]

Answer:

physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph)[1] at the surface. ... [2] Speeds higher than escape velocity have a positive speed at infinity.

6 0

2 years ago

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Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound

Roman55 [17]

Answer:

a.96cm

b.77cm

Explanation:

Highest intensity, $d_1$ =19cm

Lowest intensity, $d_2$ =29cm

$d_1+d_2=19+29=48cm$

48cm is the distance for half-wave, wavelength =48cm*2=96cm

b. From a, above we know that half-wavelength is 48cm

Also, the distance of known maximum is 29cm

Next maximum is calculated as:

$Next \m\aximum=Last \ Maximum+ \ 0.5Wavelength\\=29cm+48cm\\=77cm$

The next maximum will occur at 77cm

6 0

2 years ago

A copper wire has a radius of 2.9 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

REY [17]

Answer:

550000000N/m

Explanation:

Given that a copper wire has a radius of 2.9 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends of the wire, the wire stretches by 5.0×10−3 of its original length.

Original length L = 0.005L

the strain = extension/ original length

the strain = 0.005L / L

the strain = 0.005

Young modulus = stress / strain

11 × 10^10 = stress / 0.005

Cross multiply

Stress = 11 × 10^10 × 0.005

Stress = 550 000000 N/m

Therefore, the tensile stress on the wire is 550000000 N/m.

7 0

2 years ago

Light passes through a 0.15 mm-wide slit and forms a diffraction pattern on a screen 1.25 m behind the slit. The width of the ce

Elena L [17]

Answer:

The value is $\lambda = 900 \ nm$

Explanation:

From the question we are told that

The width of the slit is $a = 0.15 \ mm = 0.00015 \ m$

The distance of the screen from the slit is D = 1.25 m

The width of the central maximum is $y = 0.75 \ cm = 0.0075 \ m$

Generally the width of the central maximum is mathematically represented as

$y = \frac{m * D * \lambda}{a}$

Here m is the order of the fringe and given that we are considering the central maximum, the order will be m = 1 because the with of the central maximum separate's the and first maxima

So

$\lambda = \frac{a y}{ m * D }$