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BlackZzzverrR [31]
3 years ago
5

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

's acceleration is 760 m/s2 and the vertical or y component of its acceleration is 770 m/s2. The ball's mass is 0.40 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Physics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

Magnitude of net force will be 432.758 N  

Explanation:

We have given x component of acceleration a_x=760m/sec^2

And vertical component of acceleration  a_y=770m/sec^2

Mass of the ball m = 0.40 kg

So net acceleration a=\sqrt{a_x^2+a_y^2}=\sqrt{760^2+770^2}=1081.896m/sec^2

Now according to second law of motion

Force = mass × acceleration

So F = 0.40×1081.896 = 432.758 N

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Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
3 a There is a thin layer of water between the blade and the ice. Suggest how this affects friction .​
Gelneren [198K]

Answer:

The water acts like a lubricant therefore has a smooth motion over the ice.

8 0
2 years ago
If mass of both the objects are doubled
Fofino [41]

Answer:

it should be four times

4 0
2 years ago
4.
xxTIMURxx [149]
Idk what to say to this
3 0
2 years ago
A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a
zloy xaker [14]

Answer:

v = 1.4  m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block   is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5  x v² =  9.8 x .1

=  v² = 1.96

v = 1.4

v = 1.4  m /s

6 0
2 years ago
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