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3 years ago

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an

Physics

2 answers:

fredd [130]3 years ago

5 0

Answer:

520 N

Explanation:

Andreas93 [3]3 years ago

3 0

Answer:

170 N

Explanation:

Given in the question that, work a bulldozer can do = 4500 J

We will use trigonometry identity to find the distance bulldozer will travel up the hill

sin(35) = opp/hypo

sin(35) = 15/hypo

hypo = 15/sin(35)

hypo = 26.15m

Formula to use

work done = force × distance

Plug values in the above formula

4500 = force x 26.15

force = 4500/26.15

force = 172.08

force ≈ 170 N

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The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average,

3241004551 [841]

Answer:

For Pentium 4 Prescott:

% of Static Power = 10

For core i5 Ivy Bridge:

% of Static Power = 43

Given Information:

Static Power of P4 = 10 W

Dynamic Power of P4 = 90 W

Static Power of i5 = 30 W

Dynamic Power of i5 = 40 W

Required Information:

% of static power w.r.t total power dissipation = ?

Explanation:

For Pentium 4 Prescott:

% of static power = static power/total power * 100

% of static power = 10/(10 + 90) * 100

% of static power = 10/(100) * 100

% of static power = 10

For core i5 Ivy Bridge:

% of static power = static power/total power * 100

% of static power = 30/(30 + 40) * 100

% of static power = 30/(70) * 100

% of static power = 43 (rounded to nearest whole integer)

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3 years ago

White light is an _____ mixture of all the colors.

Luba_88 [7]

Answer:

White light is a combination of all colors in the color spectrum.

Explanation:

It has all the colors of the rainbow. Combining primary colors of light like red, blue, and green creates secondary colors: yellow, cyan, and magenta

6 0

3 years ago

Why does polarity allow water to be such a good solvent

Alexeev081 [22]

Water is capable of dissolving a variety of different substances, and is attracted to many other different molecules

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3 years ago

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

Ket [755]

Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

6 0

3 years ago

A person who weighs 69kg rides with an acceleration of +1.45 m/s2. What is the force exerted by the legs on the person's upper b

trapecia [35]

Answer:

The force exerted by the legs on the person's upper body 32.106 N.

Given:

Mass = 69 kg

Acceleration = 1.45 $\frac{m}{s^{2} }$

To find:

Force exerted by legs on the person's upper body = ?

Formula used:

Force = mass × acceleration

Solution:

Mass of legs = $\frac{32}{100}$ mass of body

Mass of legs = 22.08 kg

According to Newton's second law of motion,

Force = mass × acceleration

Force = 22.08 × 1.45

Force = 32.106 N

The force exerted by the legs on the person's upper body 32.106 N.

4 0

4 years ago