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3 years ago

What is weight?

Physics

2 answers:

Verizon [17]3 years ago

7 0

Answer:

option A is the right answer

hope it helps

Over [174]3 years ago

4 0

Answer:

Explanation:

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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running

Katena32 [7]

Answer:

$t_a = 13.49 \ s$

The distance is $D = 55.2 \ m$

Explanation:

From the question we are told that

The maximum speed of the cheetah is $v = 103 \ km/h = 28.61 \ m/s$

The maximum of gazelle is $u = 76.5 \ km/h = 21.25 \ m/s$

The distance ahead is $d = 99.3 \ m$

Let $t_a$ denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

$v* t_a = d + u* t_a$

=> $28.61 t_a = 99.3 + 21.25t_a$

=> $7.36 t_a = 99.3$

=> $t_a = 13.49 \ s$

Now at t = 7.5 s

$7.5 v = D+ 7.5u$

=> $28.61 * 7.5 = D + 21.25* 7.5$

=> $7.36 * 7.5 =D$

=> $D = 55.2 \ m$

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0

3 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0

3 years ago

In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a

IrinaK [193]

A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2

B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

= 224,175N

7 0

3 years ago

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

mixer [17]

Answer:

The kinetic energy of the ejected electrons increases.

Explanation:

As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.

Using conservation of energy principle we have

$E_{incident}=h\nu +\frac{1}{2}mv^{2}$

If we increase the intensity of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since $h\times \nu$ remains constant.

5 0

3 years ago

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