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2 years ago

PLEASE HELP I WILL GIVE BRALINEST

Physics

1 answer:

Greeley [361]2 years ago

6 0

Weight: it is the gravitational pull applied on a body by the earth’s surface
Mass: it is the amount of matter present

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A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0

3 years ago

Near the equator, the Earth's magnetic field points almost horizontally to the north and has magnitude B=.5 x 10^-4T. What shoul

Nataly [62]

Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>

6 0

3 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

7 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

Ahat [919]

Answer:

$V_{3.01}=-93.2m/s$

$V_{3.005}=-93.1m/s$

$V_{3.002}=-93.04m/s$

$V_{3.001}=-93.02m/s$

$V_{3}=-93m/s$

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

$Y_{3}=-99m/s$

$Y_{3.01}=-99.932m/s$

$Y_{3.005}=-99.4655m/s$

$Y_{3.002}=-99.18608m/s$

$Y_{3.001}=-99.09302m/s$

Replacing these values into the formula for average velocity:

$V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s$

$V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s$

$V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s$

$V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s$

To know the actual velocity, we derive the position and we get:

$V=27-40t = -93m/s$

5 0

3 years ago

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an

8090 [49]

Answer:

The distance is $d = 1.747 *10^{-6} \ m$

Explanation:

From the question we are told that

The order of maximum diffraction is m = 2

The wavelength is $\lambda = 775 nm = 775 * 10^{-9} \ m$

The angle is $\theta = 62.5^o$

Generally the condition for constructive interference for diffraction grating is mathematically represented as

$dsin \theta = m * \lambda$

where d is the distance between the lines on a diffraction grating

$d = \frac{m * \lambda }{sin (\theta )}$

substituting values

$d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}$

$d = 1.747 *10^{-6} \ m$

4 0

2 years ago