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3 years ago

If a cheetah goes from 10 m/s to 30 m/s in 5 seconds, what was her acceleration?

Physics

2 answers:

grandymaker [24]3 years ago

5 0

Explanation:

Y CHETOS XDXDXX

CDXCVHH

Ierofanga [76]3 years ago

3 0

Yea i i didn’t see her last name but yet bbbsnsnsnsnddx i was talking abt it to see if he could see her what she was going on with lol but i yea i she said she was like he didn’t do that too but he said it ain’t no big do but it was a weird thing

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A 4.3 kg steel ball and 6.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point

Alekssandra [29.7K]

Answer

given,

mass of steel ball, M = 4.3 kg

length of the chord, L = 6.5 m

mass of the block, m = 4.3 Kg

coefficient of friction, μ = 0.9

acceleration due to gravity, g = 9.81 m/s²

here the potential energy of the bob is converted into kinetic energy

$m g L = \dfrac{1}{2} mv^2$

$v= \sqrt{2gL}$

$v= \sqrt{2\times 9.8\times 6.5}$

v = 11.29 m/s

As the collision is elastic the velocity of the block is same as that of bob.

now,

work done by the friction force = kinetic energy of the block

$f . d = \dfrac{1}{2} mv^2$

$\mu m g. d = \dfrac{1}{2} mv^2$

$d=\dfrac{v^2}{2\mu g}$

$d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}$

d = 7.23 m

the distance traveled by the block will be equal to 7.23 m.

8 0

3 years ago

A comet is a

Juli2301 [7.4K]

The answer is D. Small object made of ice and dust that orbits the Sun and forms a coma as it approaches the Sun.

7 0

3 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

Will brainlist 20 points

Len [333]

Your answer for this question is the third option.

3 0

2 years ago

Read 2 more answers

Can you make the work output of a machine greater than the work input?

drek231 [11]

Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.

4 0

3 years ago