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dmitriy555 [2]
3 years ago
7

What is the value of x in the expression: x = p(p+q) + 4 if p = -2 and q = 7

Physics
1 answer:
Troyanec [42]3 years ago
3 0
-2(-2+7) +4 is how you would set it up


(4+(-14) + 4

-10 + 4 = -6

x = -6


The final answer you should get is -6
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List out any 4 objectives of education​
e-lub [12.9K]
- vocational aim
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6 0
2 years ago
A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running
Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

   The maximum speed of the cheetah is  v =  103 \  km/h =  28.61 \  m/s

    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0
3 years ago
In which part of the scientific method do you sum up your results
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Data Analysis and Conclusion
8 0
3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
What is relative velocity?​
joja [24]
Search it up bruhhhhhhhhhhhhh
5 0
3 years ago
Read 2 more answers
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