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fredd [130]
3 years ago
7

Edson exercises in sets that are each t minutes long. He does 6 sets of push-ups, 3 sets of pull-ups, and 4 sets of sit-ups. Use

an expression to represent the time it takes Edson to exercise as the sum of three different terms. Simplify the expression. Enter your answer in the box. The time it takes Edison to exercise is equal to t.
Physics
2 answers:
fiasKO [112]3 years ago
4 0

edson takes 130 minutes of exersise



step by step answer:

10 x 13 = 130

Nitella [24]3 years ago
3 0

Answer:

<h2>The expression would be 13t.</h2>

Explanation:

We know that:

  • Each set takes <em>t </em>minutes to be done.
  • Edson does three types of exercises with different sets: 6, 3 and 4.

Basically, we have to multiply each set by <em>t</em> minutes.

We have:

  • 6 sets of push-ups: <em> t + t + t + t + t + t = 6t</em>
  • 3 sets of pull-ups: <em>t + t + t = 3t</em>
  • 4 sets of sit-ups: <em>t + t + t + t = 4t</em>

The total amount of time considering all sets is <em>6t + 3t + 4t = 13t. </em>In other words, Edson did 13 sets, if one set takes <em>t </em>minutes, then the 13 sets takes <em>13t </em>minutes.

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GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction (\mu)=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

v^2-u^2=2as

acceleration=\mu g=0.8\times 9.8=7.84 m/s^2

s=2.4 m

0-(u)^2=2\times (-7.84)\times 2.4

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

920\times v=(920+2300)\times 6.13  (conserving momentum)

v=21.45 m/s

7 0
3 years ago
The electron drift speed in a 1.2-mm-diameter gold wire is 5.00 10-5 m/s. how long does it take 1 mole of electrons to flow thro
Vlada [557]
Twenty four minutes per second

7 0
3 years ago
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
2 years ago
In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second
lakkis [162]
If he keeps that pace he will be at the 34 yard line
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2 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
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