All categories

3 years ago

Half Time At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0

Physics

1 answer:

jenyasd209 [6]3 years ago

5 0

Answer:

3 : 08 : 10.9

Explanation:

assuming a 12 hour clock

angular velocity of the hour hand is 2π/(3600(12)) rad/s

angular velocity of the minute hand is 2π/3600 rad/s

difference is 2π/3600 - 2π/(3600(12)) = 11(2π/(3600(12)) rad/s

45° = π/4 radians

This angle is covered in a time of

π/4 rad / 22π / (3600(12)) = 900(12) / 22 = 490.909090... s

or about 8 minutes 10.9 s

ANSWER 3:08:10.9

You might be interested in

pulling one of the balls away from the other. To pull a ball away, click on the ball and drag it. What happens to the force arro

Soloha48 [4]

Gravity.: Gravity is the force that acts at a right angle to the path of an orbiting object.

7 0

3 years ago

Read 2 more answers

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se

Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0

4 years ago

What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e

Zinaida [17]

Answer:

$F=1.38*10^{-9}N$

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

$F=1.38*10^{-9}N$

5 0

3 years ago

Explain why hitting/catching a baseball are great examples of each type of collision

NeX [460]

When you hit a ball it collides with the bat. When you catch a ball it collides with your hand.

3 0

3 years ago

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

sukhopar [10]

Answer:

-6112.26 J

Explanation:

The initial kinetic energy, $KE_i$ is given by

$KE_i=0.5mv_1^{2$ } where m is the mass of a body and $v_i$ is the initial velocity

The final kinetic energy, $KE_f$ is given by

$KE_f=0.5mv_f^{2}$ where $v_f$ is the final velocity

Change in kinetic energy, $\triangle KE$ is given by

$\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})$

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for $v_f$ and 12.6 m/s for $v_i$ and 77 Kg for m we obtain

$\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J$

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>

3 0

3 years ago