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3 years ago

11

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

?

1 answer:

Arisa [49]3 years ago

4 0

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

= 1/2 (6 X 15)

= 1/2 (90)

= 45 meters

Hence, the object traveled 45 meters.

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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 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The

Temka [501]

Answer:

$1m/s^2$

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

$ma=F-f$

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

$10a=20-10=10$

$a=\frac{10}{10}=1 m/s^2$

Hence, the acceleration of the block= $1m/s^2$

4 0

2 years ago

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

$m=180 kg$ the mass of the raft

$F=75 N$ the force applied by Sawyer

$V=2 m/s$ the raft's final speed

$V_{o}=0 m/s$ the raft's initial speed (assuming it starts from rest)

We have to find the time $t$

Well, according to Newton's second law of motion we have:

$F=m.a$ (1)

Where $a$ is the acceleration, which can be expressed as:

$a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}}$ (2)

Substituting (2) in (1):

$F=m\frac{V-V_{o}}{t-t_{o}}$ (3)

Where $t_{o}=0$

Isolating $t$ from (3):

$t=\frac{m(V-V_{o})}{F}$ (4)

$t=\frac{180 kg(2 m/s-0 m/s)}{75 N}$

Finally:

$t=4.8 s$

6 0

2 years ago

Which of the following are true about S waves

shusha [124]

Answer:

3. they can travel through solids

4. they move rock at right angles to the direction of wave travel

Explanation:

S waves are called transverse waves they have the ability to move past the solids. They cannot move through the liquids, these waves are perpendicular to the direction of travel.

They are also called longitudinal waves, the ad is second to record on the seismograph as they slowly pass through the rocks. They have a speed of 3.4 to 7.2 km as per the boundary.

6 0

2 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

2 years ago