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3 years ago

Name nine elements that have been around so long that we don't even know when they

2 answers:

8 0

What was the first known element? Actually, there were nine elements known to ancient man. They were gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon.

lora16 [44]3 years ago

7 0

Answer:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

Explanation:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

- (These are the most longest elements that have been around very long )

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> The two main types of circuits are:

Gre4nikov [31]

Answer:

series and parallel

Explanation:

4 0

3 years ago

The maximum speed limit on interstate 10 is 75 miles per hour. how many meters per second is this

Dvinal [7]

Answer:

Explanation:

Given the maximum speed limit on interstate 10 as 75 miles per hour, to get the speed in meter per seconds, we need to convert the given speed to meter per seconds.

Using the conversion 1 mile = 1609.34m and 1 hour = 3600 seconds

75 miles perhour = 75miles/1 hour

75miles/1 hour (in m/s) = 75miles*1609.34m* 1 hour/1mile * 1 hour * 3600s *

= 75 *1609.34m* 1 /1 * 1 * 3600s

= 120,700.5m/3600s

= 33.53m/s

<em>Hence the maximum speed limit on interstate 10 in metre per seconds is 33.53m/s</em>

8 0

3 years ago

If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?

sweet [91]

Answer:

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6 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$