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3 years ago

A car speeds up from 8.50 m/s to 22.2 m/s in 3.84 s. what is the acceleration of the car

Physics

1 answer:

GalinKa [24]3 years ago

5 0

Answer:

3.57

Explanation:

22.2-8.50=13.7

13.7/3.84=3.57

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt.

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4 years ago

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v

Arturiano [62]

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position = 5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

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3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

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

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

4 0

4 years ago