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Bingel [31]
3 years ago
12

Una ardilla corre desarrollando una aceleracion de 3.5 m/s2 la fuerza ejecidas por las patas traseras es de 5 N. Cual sera la ma

sa de la ardilla?
Physics
1 answer:
ziro4ka [17]3 years ago
4 0

Answer:

1.43 kg

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the squirrel in the problem, we have:

a=3.5 m/s^2 is the acceleration

F = 5 N is the net force acting on it

Solving for m, we find its mass:

m=\frac{F}{a}=\frac{5}{3.5}=1.43 kg

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Sabe-se que um alqueire paulista equivale a 24200 metros quadrados. Uma chácara retangular tem um alqueire e mede 100m de frente
sattari [20]

Answer:

b = 242 m

Explanation:

A = 24200 m²

a = 100 m

b = ?

A seguinte fórmula é aplicada

A = a*b

⇒  b = A / a

⇒  b = (24200 m²) / (100 m)

⇒  b = 242 m

5 0
3 years ago
Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select
Alexeev081 [22]

Solar Energy is the answer to the question tell me if i`m wrong

7 0
3 years ago
A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.
iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2

Converting to m/s²

a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g

Hence, acceleration of the car is -0.64525g

8 0
4 years ago
A car enters a freeway with initial velocity of 15.0 m/s and with con stant rate of acceleration, reaches a velocity of 22.5 m/s
LiRa [457]

Answer:

a) The acceleration is 2.14 m/s^{2}

b) The distance traveled by the car is 65.61 m

c) The average velocity is 18.75 m/s

Explanation:

Using the equations that describe an uniformly accelerated motion:

a) a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}

b) d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m

c) v_m =\frac{d}{t}=\frac{65.61}{3.5}  =18.75 m/s

8 0
4 years ago
Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h
PolarNik [594]

Answer:

(a) a = - 201.8 m/s²

(b) s = 197.77 m

Explanation:

(a)

The acceleration can be found by using 1st equation of motion:

Vf = Vi + at

a = (Vf - Vi)/t

where,

a = acceleration = ?

Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)

Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s

t = time = 1.4 s

Therefore,

a = (0 m/s - 282.53 m/s)/1.4 s

<u>a = - 201.8 m/s²</u>

<u></u>

(b)

For the distance traveled, we can use 2nd equation of motion:

s = Vi t + (0.5)at²

where,

s = distance traveled = ?

Therefore,

s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²

s = 395.54 m - 197.77 m

<u>s = 197.77 m</u>

6 0
3 years ago
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