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3 years ago

6

A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. If friction is negligible, what is the resu

ltant force on the sled?

1 answer:

slega [8]3 years ago

8 0

Answer:

Resultant force = 17.02 N

Explanation:

As we assume the coefficient of friction is negligible, the normal force won't affect the resultant force.

As a result of this, the sine of the angle times the force of gravity on the 10 kg mass is equal to the resultant force, which is, force due to gravity = m × a = 10 kg × 9.8 m/s² = 98 N

98 sin(10)=?

Sin(10)= 0.1736

98 x 0.1736= 17.02 N.

Therefore, the resultant force = 17.02 N

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Jane is sliding down a slide. What kind of motion is she demonstrating? A. translational motion B. rotational motion C. vibratio

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A. translational motion

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3 years ago

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

3 years ago

How much heat is needed to raise the temperature of 10g of water by 16°C?

Nina [5.8K]

26°F

.............................................................

5 0

3 years ago

As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat

DaniilM [7]

Answer:

the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

Explanation:

Given that :

$\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}= \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399$

$V = e^{kt} + 399$

When v = 300 ; $\frac{dv}{dt}= - 15$

then

$\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05$

∴ $\\V = e^{0.05t} + 399\\$

Therefore, the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

6 0

3 years ago

A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?

Olenka [21]

Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
$F = k*x$

Solving:
$F = k*x$
$F = 50*0.15$
$\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark$

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
$E = \frac{k*x^2}{2}$

Solving:(Energy associated with this stretching)
$E = \frac{k*x^2}{2}$
$E = \frac{50*0.15^2}{2}$
$E = \frac{50*0.0225}{2}$
$E = \frac{1.125}{2}$
$\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago