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Doss [256]
3 years ago
5

One jet is flying east at 880 km/h,and another is traveling north at 880 km/h.Do they have the same velocity?the same speed?Expl

ain
Physics
1 answer:
shusha [124]3 years ago
6 0
Velocity is the speed and direction combined.
So, the two jets both are going at the same speed.
But they are going in different directions, so their velocities are different.
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two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f
Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

E = 1.44 \times 10^7 N/C

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

E = 2\frac{kq}{r^2}

now we have

q = 2\times 10^{-6} C

r = 5 cm = 0.05 m

now we have

E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}

E = 1.44 \times 10^7 N/C

6 0
3 years ago
Objects appear different in size and shape in a container of water due to
o-na [289]

Answer:

A. refraction of light waves

Explanation:

Refraction happens when light travels from one medium to another and changes speed and bends. This also causes objects to look different sizes and shapes when they are submerged in water.

5 0
3 years ago
Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)
olganol [36]
Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
8 0
4 years ago
Can you solve it descriptively . thanks
Solnce55 [7]

Answer:

|M_y| = 170.82 \ N.mm

Explanation:

From the diagram affixed below completes the question

Now from the diagram; We need to resolve the force at point  A into (3) components ; i.e x.y. & z directions which are equivalent to F_x \ , F_y \ ,  F_z

So;

F_x = positive x axis

F_y = Negative y axis

F_z = positive z axis

Then;

|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ?  ---- equation (3)

From equation (1); Let's make F_y the subject of the formula ; then :

F_y = \frac{77+11F_z}{27}

Substituting  the value for F_y into equation (2) ; we have:

(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since  \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm

4 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
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