Who can help me for this question plz:(

Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal

FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting 10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

$a=\frac {30}{4.077471967}=7.3575 m/s^{2}$

and T=a hence

T is approximately 7.4 N

5 0

3 years ago

Read 2 more answers

A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving

Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

$m_av_a + m_sv_s = 0$

Where $m_a = 92kg$ is the mass of the astronaut, $m_s = 1200kg$ is the mass of the satellite, $v_s = 0.14 m/s$ is the speed of the satellite. We can calculate the speed $v_a$ of the astronaut:

$v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s$

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0

3 years ago

To be effective, an exercise program must be?

ExtremeBDS [4]

To be effective, an exercise program must have an aerobic form, portion for strength enhancement, and a stretching part. These three things are essential because they each target specific improvements in your body. For example, aerobics can help you maintain your body’s fitness or make it better. This usually targets your heart rate and ensures that you burn fat while doing so. Second is strength enhancement; this will make sure that your body becomes better – not just in a feeble state. Lastly is stretching, your muscles are like rubber bands. You cannot end or start your exercise program without stretching simply because they can damage your muscles as well. Aside from this, stretching can stop you from shocking your body into a physical activity, which may cause you to lose consciousness or have undue stress and fatigue.

3 0

3 years ago

Read 2 more answers

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0

3 years ago

A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha

Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

$N = \dfrac{m g}{cos\theta-\mu sin \theta}$

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

$\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}$

taking cos θ from nominator and denominator

$F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g$

$\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}$

now, inserting all the given values

$\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}$

μ = 0.6

7 0

3 years ago