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RUDIKE [14]
3 years ago
13

You arrive in my class 45 seconds after leaving math which is 85 meters away. How fast did you travel?

Physics
1 answer:
levacccp [35]3 years ago
7 0
Speed = 85/45
units m/s
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A rock drops from a height of 60 m. how long does it take for it to hit the ground
Svet_ta [14]
Here, Apply 2nd equation of the Kinematics :
S = ut + 1/2 at²
Here, s = 60 m
u = 0   [ free fall ]
a = 9.8 m/s²  [ constant value for Earth system ]

Substitute their values, 
60 = 0*t + 1/2 * 9.8 * t²
120 = 9.8t²
t² = 120 / 9.8
t = √12.24
t = 3.50 s

In short, Your Answer would be 3.50 seconds

Hope this helps!
7 0
3 years ago
if a runner's is 400 w as she runs,how much chemical energy does she convert into other forms in 10.0 minutes?
Fiesta28 [93]

Explanation:

The runner runs for a total time of

\Delta t=10.0 min= 600 sΔt=10.0min=600s

The energy converted by the runner during this time is equal to the power of the runner times the total time:

E=P \Delta t=400 W \cdot 600 s =2.4 \cdot 10^5 JE=PΔt=400W⋅600s=2.4⋅10

5

J

7 0
4 years ago
The gravitational force between a satellite and Earth’s moon is 324 N. The mass of the moon is 7.3 × 1022 kg.
tiny-mole [99]

450 kg is your answer middle school physic


8 0
3 years ago
Read 2 more answers
A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T
Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

  a) The minimum force of the wind that turns the system is Fw = 17.64 N

  b) The system resists much greater forces because the base has more mass

 Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

               Σ τ = 0

Where τ is the torque  

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

         cos 15 = \frac{y_w}{2.35}

         y_w = 2.35 cos 15

Post Weight

        sin 15 = \frac{x_p}{2.00}

         xp = 2.0 sin 15

Base weight

         cos (90-15) = \frac{x_b}{0.25}

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          F_w \ y_w  + W_p \ x_p - W_b \ x_b = 0

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is W_p = 26.0 N and the weight of the base with water is W_b = 810 N

     F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}

     F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}

       

Let's  calculate

     F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}

     F_w = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

  a) the minimum force of the wind that turns the system is Fw = 17.64 N

  b) the system resists much greater forces because the base has more mass

Learn more  here: brainly.com/question/7031958

3 0
3 years ago
SCIENCE HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Andrews [41]
It's by..................
energy/time
6 0
4 years ago
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