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2 years ago

Calculate the weight of a 4.5 kg rabbit.

Physics

1 answer:

solniwko [45]2 years ago

3 0

The correct answer is: 13900589.

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A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast

Neko [114]

Answer:

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of $\frac{1}{12}$ × 360 = 30m

And turtle covers a distance of $\frac{11}{12}$ × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

8 0

2 years ago

Read 2 more answers

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

2 years ago

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

$r^2=\dfrac{2\times 2\times 1.1}{(5)}$

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0

2 years ago

Read 2 more answers

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$