Which energy transformation converts energy from the sun’s core to light energy needed by plants?

What is the distance moved by a horse galloping at 75 km/h in 10.0 s?

skelet666 [1.2K]

125Answer:

Explanation:uruhavdfh

4 0

3 years ago

People often think that Galileo dropped two objects of dramatically different mass off of the Leaning Tower of Pisa that both hi

kramer

Answer: because of air resistance. See explanation for further details.

Explanation: Galileo performed an experiment to proof that the time of descent of two different masses is independent of time.

But in reality this is most likely not true because of air resistance and other fluid frictional effects in consideration.

If the experiment is performed in a vacuum, it will always be true that time is independent of masses of two falling objects.

3 0

3 years ago

A car is driven 225 km west and then 98 km southwest (45).

slava [35]

Sum the vector components: Dx = 225* Cos(180)+ 78*Cos(225)= -280.154 km Dy = 225* Sin(180)+ 78*Sin(225) = -55.154 km displacement: Sqrt(Dx^2+Dy^2) = 285.532 km Arctan(Dy/Dx) = 191.137degrees CCW OR: 11.137 degrees South of West

4 0

3 years ago

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A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti

vovikov84 [41]

Answer:

<u> $\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$ </u>

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g = 10 × 9.8 = 98 N

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied

10 kg block will slide on 40 kg slab and net force on it

= 100 N - kinetic friction

$=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)$

= 100 - 39.2

= 60.8 N

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}$

$\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }$

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4

Frictional force on 40 kg slab by 10 kg block = 39.2 N

$40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}$

$40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}$

40 kg slab will move with = $0.98 \mathrm{m} / \mathrm{s}^{2}$

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to

natali 33 [55]

Answer:

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Magnitude of resulting torque = 3.84 Nm
Angular acceleration = = 0.0086877 rad/s² rad/s²

Explanation:

It is given that:

I = 442 kg˙m2

rx = 0.76 m, ry = 0.035 m, rz = 0.015 m,

Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N

F = Fx i + Fy j + Fz (equation 1)

r = rx i + ry j + rz k (equation 2)

(a) Torque

T = r * F (equation 3)

by putting equation 1 and 2 into equation 3, we have;

Torque= r x F

= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )

= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

Therefore,

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

b) Magnitude of the torque

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

=(0.025 x 2.2 - (0.035 x (-1.2))) i +(0.035 x (-7.8)) - 2.19 x 2.2) j+(2.19 x 1.2 - 0.025 - (-7.8)) k

= (0.055 - 0.042) i + (0.273 - 4.818) j + (-2.628 -0.195) k

= (0.013) i - (-4.545) j + (-2.433) k

Magnitude of resulting torque = √0.013² + 4.545² - 2.433²

=√14.751

= 3.84 Nm

c) Angular acceleration

Since

angular acceleration = torque/moment of inertia

= 3.84/ 442

= 0.0086877 rad/s²

Hence,

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Magnitude of resulting torque = 3.84 Nm
Angular acceleration = = 0.0086877 rad/s² rad/s²

To learn more about Torque, use the given link:

brainly.com/question/19247046?referrer=searchResults

#SPJ4

3 0

1 year ago

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