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2 years ago

6

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what are the X AND Y components of the NORMAL FO

RCE?

1 answer:

Lera25 [3.4K]2 years ago

4 0

Answer:

y-component = 45.75N

x-component = 0N

Explanation:

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An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the o

slavikrds [6]

Answer:

Final velocity of the object(v) = 114 m/s

Explanation:

Given:

Initial velocity (u) = 14 m/s

Accelerates (a) = 5 m/s²

Time taken = 20 seconds

Find:

Final velocity of the object(v) = ?

Computation:

According to 1st law of motion.

⇒ v = u +at

⇒ v = 14 + (5)(20)

⇒ v = 14 + 100

⇒ v = 114 m/s

Final velocity of the object(v) = 114 m/s

8 0

3 years ago

If you start rolling down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill ypu KE

Serjik [45]

Answer:

$v=\sqrt{(2x/m)}\ \ m/s$

Explanation:

Potential Energy= Kinetic Energy

Let $x$ be the value of Kinetic Energy.

We know that

$PE=KE=x\\x=\frac{1}{2}mv^2\\$

Make $v$ the subject of the formula to get speed at the bottom of the hill.

$v^2=2x/m\\v=\sqrt{2x/m}$

8 0

3 years ago

Squids are the fastest marine invertebrates, using a powerful set of muscles to take in and then eject water in a form of jet pr

tatuchka [14]

Answer:

0.25 m/s

Explanation:

This problem can be solved by using the law of conservation of momentum - the total momentum of the squid-water system must be conserved.

Initially, the squid and the water are at rest, so the total momentum is zero:

$p_i = 0$

After the squid ejects the water, the total momentum is

$p_f = m_s v_s + m_w v_w$

where

$m_s = 1.60 kg$ is the mass of the squid

$v_s$ is the velocity of the squid

$m_2 = 0.115 kg$ is the mass of the water

$v_w = 3.50 m/s$ is the velocity of the water

Due to the conservation of momentum,

$p_i = p_f$

so

$0=m_s v_s + m_w v_w$

so we can find the final velocity of the squid:

$v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.115 kg)(3.50 m/s)}{1.60 kg}=-0.25 m/s$

and the negative sign means the direction is opposite to that of the water.

8 0

3 years ago

The resistanceless inductor is connected across the ac source whose voltage amplitude is 24.5 V and angular frequency is 850 rad

timurjin [86]

Answer:

Part A 2.88 A, PART B 28.82 mA PART C 0.288 mA

Explanation:

We have given angular frequency $\omega =850rad/sec$

Voltage source has an amplitude of 24.5 V, So V= 24.5 volt

Part A

We have given inductance $L=10^{-2}H$

So inductive reactance $X_L=\omega L=850\times 10^{-2}=8.5ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A$

Part B

We have given inductance $L=1 H$

So inductive reactance $X_L=\omega L=850\times 1=850ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA$

Part C

We have given inductance $L=100 H$

So inductive reactance $X_L=\omega L=850\times 100=85000ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA$

8 0

3 years ago

A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Tems11 [23]

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet

The wooden block is initial at rest $(V_{wi} = 0)$ this yields

$m_{b}V{b_{i} } = m_{w} V_{wf} + m_{b}V_{bf}$

By solving for $V_{wf}$ adn substitute the givens

$V_{wf}$ = $\frac{m_{b}V_{bi} - m_{b} V_{bf} }{m_{W} }$

= $\frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}$

$V_{wf}$ = 4.67m/s

b) The center of mass speed is defined as

$V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}$

substituting:

$V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)$

V = 8.29 m/s

7 0

3 years ago