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evablogger [386]

2 years ago

7

What is the momentum of and 200 kg car traveling south at 22 m/sec?

1 answer:

motikmotik2 years ago

5 0

The momentum of the car is 4.4x10^3 kg•m/sec

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Please, what is a car battery pH?

Mama L [17]

Answer

about – 0.7

Explanation:

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3 years ago

A risk-free, zero-coupon bond with a $5000 face value has ten years to maturity. The bond currently trades at $3650. What is the

PtichkaEL [24]

Answer:

The Yield to Maturity of the bond is YTM = 3.20%

Explanation:

Mathematically the Yield to Maturity of the bond YTM is as follows

$YTM = \frac{C + \frac{F-P}{n} }{\frac{F+P}{2} }$

Where C is the amount of payment to be made = $0

P is the price i.e the present value =$3650

F is the face value of the bond=$5000

n is the year of maturity of the bond = 10 years

$YTM = \frac{0+\frac{5000-3650}{10} }{\frac{5000+3650}{2} } * \frac{100}{1}$

$=3.20$ %

5 0

3 years ago

Why is the electromagnetic spectrum called a spectrum?

Doss [256]

Answer & Explanation:

Scientists call them all electromagnetic radiation. The waves of energy are called electromagnetic (EM) because they have oscillating electric and magnetic fields. Scientists classify them by their frequency or wavelength, going from high to low frequency (short to long wavelength).

6 0

3 years ago

Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no

Cloud [144]

Answer:

Propels in the opposite direction

Explanation:

7 0

2 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago