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Marysya12 [62]
3 years ago
12

Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd

ed goalpost and experiences a backward force of 17700 N for 5.50×10−2 s
Physics
1 answer:
Free_Kalibri [48]3 years ago
5 0

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
lubasha [3.4K]

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

6 0
3 years ago
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0
3 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

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