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Verdich [7]
3 years ago
13

1. Define weight

Physics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

  1. weight is the product of mass andgravitational acceleration expressed in newtons.
  2. w=mg
  3. w=mg ....w=50kg X 9.8m/s...w=490N
You might be interested in
What is the difference between reflection and refraction
sineoko [7]

Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

Explanation:

I just learned about this 2 weeks ago actually.

5 0
3 years ago
A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40
Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is \frac{1}{2}mv^2

So total energy at the top of the hill =mgh+0=mgh

At the bottom of the hill kinetic energy is equal to \frac{1}{2}mv^2 and potential energy will be 0

So total energy at the bottom of the hill is equal to 0+\frac{1}{2}mv^2

Form energy conservation mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}, for v = 20 m/sec

20=\sqrt{2\times 9.8\times h}

Squaring both side

19.6h=400

h = 20.408 m

Now if velocity is 0 m/sec

40=\sqrt{2gh}

19.6h=1600

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
4 years ago
What does it mean by the fact that the acceleration of a car is 5m/s square??​
dalvyx [7]

Answer:

An acceleration of 5m/s^2 means that the velocity of a body is increasing by 5m/s per second in a certain direction

Explanation:

5 0
3 years ago
A 6.00-kg box is sliding to the right across the horizontal floor of an elevator. The coefficient of kinetic friction between th
kap26 [50]

Answer: Kinetic frictional force = 23.76N

Explanation: Please see the attachments below

7 0
3 years ago
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