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3 years ago

Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración

Physics

1 answer:

guajiro [1.7K]3 years ago

6 0

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

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A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return

tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

7 0

3 years ago

A projectile is fired vertically from Earth's surface with

scoray [572]

Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let m = Mass of the projectile

Solution:

Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

$-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2$

$-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )$

= $( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$

5 0

3 years ago

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is

AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

$W'=W-B$

which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

$\rho = \frac{m}{V}$

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

3 0

3 years ago

Find :

Pepsi [2]

Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

a) h = v₀t + ½gt² = 40(4) + ½(-10)4² = 80 m

v² = u² + 2as

h = (0² - 40²) / 2(-10) = 80 m

8 0

2 years ago

4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data

melisa1 [442]

Answer:

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Explanation:

5 0

2 years ago