HELP this need to be turned it tomorrow

Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449

Anni [7]

Answer:

(a) $a\approx1.4 m.s^{-2}$

(b) $v\approx 4.133 m.s^{-1}$

Explanation:

Given:

Limiting tension of snapping of the rope, T= 387 N

Weight of the object to be lifted, $w=449 N$

∴mass, $\Rightarrow m= 45.8163 kg$

height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵ $F=m.a$

$62=45.8163\times a$

$a= \frac{62}{45.8163}$

$a\approx 1.4 m.s^{-2}$

(b)

From the equation of motion ,we have:

$v^{2} =u^{2} +2a.s$ ....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

$a= 1.4 m.s^{-2}$

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

$v^2= 0^2 + 2\times 1.4\times 6.1$

$v\approx 4.133 m.s^{-1}$ is the velocity with which the body will hit the ground in the given conditions.

7 0

4 years ago

A 0.10 kg meter stick is held perpendicular to a vertical wall by a 2.5 mm string going from the wall to the far end of the stic

il63 [147K]

Explanation:

a)

Sum of moments = 0 (Equilibrium)

T . cos (Q)*L = m*g*L/2

$cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}$

$T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2$

$T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }$

b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!

c)

$T = \frac{1.962}{\sqrt{(2)^2 - L^2} }$

6 0

3 years ago

You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an

Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0

3 years ago

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for t

egoroff_w [7]

Answer:

a. 5 batteries b. 1050 mAh

Explanation:

Here is the complete question

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?

Solution

A) How many batteries are needed?

Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries

So, the number of batteries needed is 5.

B) What mAh capacity should the batteries have?

Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.

So, Q = It = 150 mA × 7 h = 1050 mAh.

So, the batteries should have a mAh of 1050 mAh

4 0

3 years ago

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Aleksandr-060686 [28]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m