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3 years ago

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A 730 kg Mini runs into a stationary 2500 kg SUV. If the Mini was moving at 10. M , how s fast will it be moving after making a

completely inelastic collision with the SUV?

1 answer:

spayn [35]3 years ago

5 0

Answer:

2.26m/s

Explanation:

Since it is an inelastic collision, the mini and the Suv move together after the collision. The law of conservation of momentum states that the total momentum before and after the collision is conserved:

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given that,

$m_{1}$ = 730 kg

$u_{1}$ = 10 m/s

$u_{2}$ = 0

$m_2 =$ 2500kg

v = ?

When substituting the given values in the above equation, we get

$730(10) + 2500(0) = (730+2500)v\\\\7300 + 0 = 3230v\\\\v = \frac{7300}{3200} \\\\= 2.26m/s$

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A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from

Mrac [35]

Answer:

C) True. At maximum displacement, its instantaneous velocity is zero.

Explanation:

The simple harmonic movement is given by

x = A cos wt

Speed

v = - A w sin wt

At the point of maximum displacement x = A

A = A cos wt

cos wt = 1

wt = 0

We replace the speed

v = -Aw sin 0 = A w

Speed is maximum

Let's review the claims

A) False. Speed is zero

B) False. It can be determined

C) True. Agree with our result

D) False. When one is maximum the other is minimum

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3 years ago

lozanna [386]

Answer:

It’s called a conservative field.

Explanation:

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2 years ago

Read 2 more answers

A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horiz

dalvyx [7]

Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

8 0

3 years ago

Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi

Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0

3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

3 years ago