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2 years ago

What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air

Physics

1 answer:

madam [21]2 years ago

8 0

Hi there!

We can use the kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

$v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}$

Substitute in the given values:

$v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}$

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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What

Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

$a_{c} =\frac{v^{2} }{r}$

Where $a_{c}$ is the centripetal acceleration

$v$ is the velocity

and $r$ is the radius

From the given information

$v = 12 \ m/s$

and $r = 30 \ m$

Therefore,

$a_{c} =\frac{12^{2} }{30}$

$a_{c} =\frac{144 }{30}$

$a_{c} = 4.8\ m/s^{2}$

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

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2 years ago

The process of heat radiation is the ONLY method of heat transfer that can occur

lina2011 [118]

Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. so...<span> electromagnetic waves ?</span>

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3 years ago

Read 2 more answers

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

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3 years ago

a 70 kg skydiver opens her parachute. The force due to air resistance is now 1200 N. what is the acceleration of the skydiver

Natasha_Volkova [10]

Around 50 I think so

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3 years ago

An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o

borishaifa [10]

Hi there!

Recall the equation for centripetal force:
$F_c = \frac{mv^2}{r}$

We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
$r * F_c = mv^2$

Divide both sides by Fc:
$\boxed{ r= \frac{mv^2}{F_c}}$

7 0

2 years ago