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PilotLPTM [1.2K]

4 years ago

12

A stone dropped from the root of a high buliding . A second stone is dropped 1.30 s later How far apart are the stones when the

second one has reached a speed of 14.0 m/s

1 answer:

oksian1 [2.3K]4 years ago

6 0

Answer:

The stones are apart one each the other 26.37 meters when the second one has reached a speed of 14 m/s.

Explanation:

<u>Second Stone:</u>

Vb= g*t

tb (14m/s)= Vb/g

tb (14m/s)= (14 m/s) / (9.8 m/s²)

tb (14m/s)= 1.42 sec

Hb= (g * tb²)/2

Hb= 9.88m

<u>First Stone:</u>

ta= tb + 1.3 sec

ta= 2.72 sec

Ha= (g*ta²)/2

Ha= 36.25m

Distance between Stones:

Ha-Hb= 36.25m - 9.88m

Ha-Hb= 26.37m

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Find the equivalent resistance of this

frosja888 [35]

Answer:

Re=160ohm

Explanation:

Step#1

Rt=R1+R2 ( because both are in series)

Rt=(100+220 ) ohm

Rt=320 ohm

Step#2

Rt and R3 are parallel so,

Re= (Rt× R3) ÷ (Rt+R3)

Re= (320×320)÷( 320+320)

Re = 102,400÷ 640

Re=160ohm

7 0

3 years ago

Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

7 0

3 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

7 0

3 years ago

A constant net torque is exerted on an object. Which of the following quantities for the object cannot be constant? (Select all

prohojiy [21]

Answer:

A. kinetic energy

B. angular velocity

E. angular position

Explanation:

The quantities that cannot be constant if a constant net torque is exerted on an objecta are:

A. Kinetic energy. If a torque is applied, the linear or angular speed will be changing at a rate proportional to the torque, so the kinetic energy will change too.

B. Angular velocity. It will change at a rate equal to the torque.

C. Angular position. If the angular velocity changes, the angular position will change.

3 0

4 years ago

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

7 0

3 years ago

Read 2 more answers