Is this statement true or false? Most metals in the chart react with oxygen.

4. A steel cable spanning a river is 220.000 m long when the temperature is 30.°C.

romanna [79]

Answer:

219.9208 m

Explanation:

The new length is given by

New length= Original length *(1-Temperature change*coefficient of

thermal expansion of steel)

Here, the change in temperature is $30^{\circ}-0^{\circ}= 30^{\circ}$

New length= $220.000(1-30*12x10^{-6}=219.9208m$

Therefore, the new length will be 219.9208 m

5 0

3 years ago

a sleepy atudent drops a calculator out of a window yhata 20.7 m off the geound. we can ignore air resistance. what js the veloc

frez [133]

The acceleration of gravity is

9.8 m/s^2 down.

When an object falls out of a hand, its speed after 1.8s is

(9.8)x(1.8) = 17.6 m/s down.

It doesn't matter what it is, how much it weighs, or how high it was dropped from.

If it's more than 17.6 m/s, then this happened on a different, bigger planet.

If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.

6 0

2 years ago

In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mas

Aneli [31]

Answer:

a. ΔP/Δt = 42.6 N

b. F = 42.6 N

c. P = 142042.4 Pa = 1.42 KPa

Explanation:

a.

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

Vf = Final Speed = 1110 m/s

Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

ΔP/Δt = 42.6 N

b.

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

F = 42.6 N

c.

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

P = 142042.4 Pa = 1.42 KPa

4 0

3 years ago

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:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$