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3 years ago

calculate the linear momentum of a pickup truck that has a mass of 1000kg, and is traveling eastward at 20 m/s

2 answers:

ioda3 years ago

4 0

Momentum = Mass × Velocity

From the given data, Mass is 1000kg and Velocity is 20m/s

Momentum = 1000 × 20
$Momentum = 20000 {kgms}^{ - 1}$

Sloan [31]3 years ago

3 0

Answer:

The momentum of the truck is $2\times 10^4\ kg-m/s$ .

Explanation:

It is given that,

Mass of the truck, m = 1000 kg

Speed of the truck, v = 20 m/s (towards east)

Let p is the linear momentum of a truck. It is equal to the product of mass and velocity. Mathematically, it is given by :

p = m v

$p=1000\ kg\times 20\ m/s$

$p=2\times 10^4\ kg-m/s$

So, the linear momentum of the truck is $2\times 10^4\ kg-m/s$ . Hence, this is the required solution.

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FrozenT [24]

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2 years ago

the three major scales used to measure earthquakes are Mercalli scale Richard scale and magnitude scale

iVinArrow [24]

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3 0

2 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago

A resistor is connected to a 36v power supply. An ammeter measures a current of 2.0 A going through it. Determine the resistance

m_a_m_a [10]

R = 18 ohms

Explanation:

Given:

V = 36 volts

I = 2.0 A

R = ?

Use Ohm's law to solve for the resistance:

V = IR

R = V/I

= (36 volts)/(2.0 A)

= 18 ohms

8 0

2 years ago