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AfilCa [17]
3 years ago
14

A tennis ball is thrown against a vertical concrete wall that is fixed to the ground. The ball bounces off the wall. How does th

e force exerted by the ball on the wall compare with the force exerted by the wall on the ball
Physics
1 answer:
cestrela7 [59]3 years ago
8 0

Answer:

Explanation:

The forces compare together as a result of the fact that the force exerted by that of the ball and the force exerted by that of the wall both have the same magnitude.

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The magnitude of the electric force between two protons is 2.30 10-26 n. how far apart are they?
SCORPION-xisa [38]
The electric force between two charge objects is calculated through the Coulomb's law.
                               F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
                           2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m. 
7 0
3 years ago
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Please help me someone !
san4es73 [151]

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

7 0
3 years ago
A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.
Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0
3 years ago
Read 2 more answers
The momentum of a 0.1 kg object traveling at 2000 m/s is 20,000 kg·m/s. True or False
Alina [70]

That's false.  

The definition of momentum is (mass) x (speed), so they must be multiplied.

"20,000 kg-m/s" has the correct units resulting from multiplication, but the number could only be the result of division.

3 0
3 years ago
A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
soldier1979 [14.2K]

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

v_f=\frac{153.7}{40.7}=3.78 m/s

3 0
3 years ago
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