5. A wire, resistance 23.6 carries a current. 5.9 A. What is the

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour for 4 hours time. What is the total distan

dimulka [17.4K]

Answer:

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour for 4 hours time. What is the total distance traveled by the comet during this time?

✓ Distance = Speed * Time Distance = 50,000 * 4 Distance = 200,000 km. Hope this helps

Explanation:

4 0

2 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0

3 years ago

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

Dvinal [7]

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

$s=0+\0.5\times 21\times (2)^2$

$s=42\ m$

so Jet Plane travels a distance of 42 m in 2 s

5 0

3 years ago

If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d

omeli [17]

Answer:The central spot becomes Dark

Explanation: it become dark because as the wavelength reduces,the velocity in the detector decreases, this time by 90degrees

4 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago