Which of the following is contact force

A 0.0663 kg ingot of metal is heated to 241◦C

Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0

3 years ago

A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. While in the water, the ball e

denis23 [38]

The initial velocity is
v(0) = 16.5 ft/s

While in the water, the acceleration is
a(t) = 10 - 0. $\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}$

The velocity function is
$v(t)=12.5+4e^{-0.8t}$
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
$v(5.7)=12.5+4e^{-0.8\times5.7} = 12.54 \, ft/s$

The depth of the lake is
$d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft$

Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft

7 0

3 years ago

An athlete wants to refill an old punching bag used for training, as shown in the picture. When refilled, the bag will rip if it

In-s [12.5K]

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states action and reaction are equal and opposite. That is the force applied to an object is equal in magnitude to force experienced by the object but in opposite direction.

From the given question, when the bag exert a certain on the athlete, the athlete also exerts similar force to the bag but in opposite direction.

Thus, the complete sentence is as follows;

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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5 0

2 years ago

A 6.00 μFμF capacitor that is initially uncharged is connected in series with a 4.00 ΩΩ resistor and an emf source with EE EMF =

Sindrei [870]

Answer: Rr = 900w

T = 1.66×10^-5s

P = 36.92W

Explanation: please find the attached file for the solution.

3 0

3 years ago

Read 2 more answers

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.

Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s$

Maximum height

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m$

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s$

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0

3 years ago