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Tamiku [17]
3 years ago
13

10. At point A, a runner is jogging 5 m/s. Forty seconds later, at point B

Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

Deceleration of 0.075m/s²

Explanation:

Given parameters:

Initial velocity  = 5m/s

Time  = 40s

Final velocity  = 2m/s

Unknown:

The jogger's acceleration  = ?

Solution:

To solve this problem, we use;

         a  = \frac{v - u}{t}  

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

      a  = \frac{2 - 5}{40}   = - 0.075m/s²

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how might testing small-scale building models or their components be helpful in testing improvements and changes to the building
azamat

Answer: & Explanation:

small scaled building models are very rare to make/build nowadays.

due to advance technology it is more wiser and more convenient to use 3d programs like revit in creating building model to scale.

so It is always a good idea to make models of your ideas, before deciding on the final design.

This will help you determine whether your idea is going to work or needs modifying. A model allows you to test your solution quickly and cheaply.

Reworking a massive building takes years, millions of pounds, and meticulous planning and project management.

4 0
4 years ago
If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

6 0
4 years ago
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mar
xz_007 [3.2K]

Answer:

0.705 m/s²

Explanation:

a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.

Using newton's law of motion:

v² = u² + 2as

v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h,  s = distance = 67 m

9.72² = 0² + 2a(67)

134a = 94.484

a = 0.705 m/s²

b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:

v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m

v² = u² + 2as

9.72² = 9.72² + 2a(88)

176a = 9.72² - 9.72²

a = 0

c) During the last distance, the speed slows down from 35 km/h to 32 km/h.

u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m

v² = u² + 2as

8.89² = 9.72² + 2a(45)

90a = 8.89² - 9.72²

90a = -15.4463

a = -0.1716 m/s²

The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.

8 0
3 years ago
Let us suppose the magnitude of the original Coulomb force between the two charged spheres is FF. In this scenario, a third sphe
Vinil7 [7]

Answer:

F ’= 1/32 F

We see that the value of the force is the initial force over 32

Explanation:

In this problem the sphere that is touching the others is connected to ground, after each touch,

Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half

½ (Q / 2) = ¼ Q

For the red dial repeat the same scheme

with the first touch the charge is reduced to Q / 2

with the second touch e reduce to ½ (Q / 2) = ¼ Q

with the third toce it is reduced to ½ (¼ Q) = ⅛ Q

Now let's analyze what happens to the electric force

if the force is F for when the charge of each sphere is Q

        F = k Q Q / r²

with the remaining charge strength is

        F ’= k (¼ Q) (⅛ Q) / r²

        F ’= 1/32 k Q Q / r²

        F ’= 1/32 F

We see that the value of the force is the initial force over 32

6 0
4 years ago
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