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Anastasy [175]
3 years ago
7

Which word identifies a large natural or human-made lake used to supply water?

Physics
2 answers:
lutik1710 [3]3 years ago
7 0

Answer:

reservoir.     C

Explanation:

Just took the test on edge 2020

Sidana [21]3 years ago
4 0
Reservoir i think. i’m not sure
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a rectangular tank measures 12.5 metres long 10.0 wide and 2.0 metre high calculate the mass of the water in the tank when it is
Tom [10]

Answer:

250000kg

Explanation:

Mass =density * volume

Density=1000kg/m³

Volume=l*w*h

12.5*10.0*2.0=250m³

Mass=1000 *250

=250000kg

7 0
3 years ago
The greater the mass of an object, the greater its force due to gravity
Serga [27]
What is the question?
6 0
3 years ago
A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
Jlenok [28]

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

5 0
3 years ago
un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
3 years ago
While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?
timurjin [86]
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x). 
5 0
3 years ago
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