there are 1.6 km in a mile. the distance between two cities is 248 miles. How many kilometers apart are the two cities?

A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t

grandymaker [24]

Answer:

13.23J

Explanation:

PE = m*g*h

PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)

3 0

2 years ago

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A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light

MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

v = d / t

t = d / v

t = d n / c

Let's look for the times on each sheet

Ice

t₁ = 1.4 10⁻² 1.31 / 3 10⁸

t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

t = t₁ + t₂

t = (0.6113 + 1.52) 10⁻¹⁰

t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

d = c t

d = 3 10⁸ 2.13 10⁻¹⁰

d = 6.39 10⁻² m

d = 6.39 cm

3 0

2 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

3 0

3 years ago

A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$