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3 years ago

A cadillac escalade has a mass of 2 569.6 kg, if it accelerate at 4.65m/s2 what is the net force of the car

1 answer:

3 0

Force is equal to mass multiplied by acceleration, therefore
F=ma
m=2569.6 kg
a=4.65m/s^2
therefore F=2569.6*4.65=11948.6 (correct to 1 d.p.)

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What is the pressure exerted when a force of 20 N is applied to an object over an area of 2 m2 ?

IgorC [24]

(20 N)/(2 m²) = (20/2) N/m² = 10 Pa

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The pascal (Pa) is the derived SI unit of pressure equal to 1 N/m².

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3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago

Sally Sizzle builds a circuit by connecting three resistors in parallel with a 9V battery. The current in this circuit is 3.6 am

Andru [333]

Then answer is C 9 ohms

7 0

3 years ago

Read 2 more answers

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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3 years ago

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3 years ago