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3 years ago

A example of using a muscular strength?

Physics

2 answers:

DIA [1.3K]3 years ago

7 0

An example of using muscular strength would be doing things like lifting weights and things like that.

Alika [10]3 years ago

5 0

An example of using muscular strength would be lifting a heavy object, such as a weight or gallon of milk.

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11) Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a sp

sleet_krkn [62]

Answer:

100 m

Explanation:

Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t

Speed of Arthur = 3 m/s

Speed of Betty = 2 m/s

Distance = Speed × time

Distance covered by Arthur = 3t

Distance covered by Betty = 2t

The distance covered by both of them will be 100 m

$3t+2t=100\\\Rightarrow 5t=100\\\Rightarrow t=\frac{100}{5}\\\Rightarrow t=20\ s$

The speed of dog is 5 m/s

Spot is running back and forth at 5 m/s for 20 seconds

In 20 seconds the distance covered by the dog is

$\mathbf{5\times 20}=\mathbf{100\ m}$

5 0

4 years ago

A 4-N object object swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the string is

Amanda [17]

Answer:

Explanation:

Given mg = 4N .

m = 4 / g

At the bottom of the swing let centripetal acceleration be a

T - mg = ma

9 - 4 = ma

5 = 4 a / g

a = 5g / 4

6 0

3 years ago

uring the investigation of a traffic accident, police find skid marks 89.9 m long. They determine the coefficient of friction be

aivan3 [116]

Answer:

$V_{0}=29.68m/s$

Explanation:

In order to solve this problem, we must first do a drawing of the situation (see attached picture).

When the brakes of the car were applied, we can see that there was only one horizontal force affecting the vehicle's movement, which was the force of friction. When analyzing the free body diagram we can apply Newton's laws to determine the equations we will use to solve this.

$\sum F_{x}=ma$

so:

-f=ma

we also know that:

$f=N\mu_{k}$

$-N\mu_{k}=ma$

we can now solve for the acceleration, so we get:

$a=\frac{-N\mu_{k}}{m}$

we don't know what the normal force is, so we can find it out by analyzing the vertical forces applied to the car:

$\sum F_{y}=0$

so we get:

N-W=0

which means that:

N=W

we also know that:

W=mg

N=mg

we can substitute this into the first equation so we get:

$a=\frac{-mg\mu_{k}}{m}$

which simplifies to:

$a=-g\mu_{k}$

we can now substitute the provided data:

$a=(-9.8m/s^{2})(0.5)$

which yields:

$a=-4.9m/s^{2}$

once we got the acceleration, we can use kinematics formulas to solve this, we got the following formula:

$a=\frac{V_{f}^{2}-V_{0}^{2}}{2x}$

we know the final velocity must be zero, since that's where the car got to a stop, so the formula then becomes:

$a=-\frac{V_{0}^{2}}{2x}$

we can now solve for the initial velocity, which yields:

$V_{0}=\sqrt{-2xa}$

so we can now substitute the daa we know, so we get:

$V_{0}=\sqrt{-2(89.9m)(-4.9m/s^{2})}$

so we get:

$V_{0}=29.68m/s$

So from this we know that the velocity of the car must have been of at least 29.68m/s when the brakes were applied.

4 0

4 years ago

Can you please help me with this physics question

erastova [34]

Answer:

See the answers below

Explanation:

We can solve both problems using vector sum.

Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative

$F_{resultant}=100+30-85+900\\F_{resultant}=845[N]$

The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.

$F_{drag}=50[N]$

Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.

$F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN]$

6 0

3 years ago

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

8 0

4 years ago