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3 years ago

9

____________ is an individual sport that helps develop your hand-eye coordination.

1 answer:

Snezhnost [94]3 years ago

4 0

Answer:

Answer option A) Table Tennis helps develop your hand-eye coordination.

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A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

Elanso [62]

$F=55 \ N \\ d=10 \ m \\ \boxed{W-?} \\ \bold{Solving:} \\ \boxed{W=F \cdot d} \\ W=55\ N \cdot 10 \ m \\ \Rightarrow \boxed{W=550 \ J}$

6 0

3 years ago

Read 2 more answers

Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel

krek1111 [17]

Answer:

The correct answer is <u>option (A) that is KEA > KEB .</u>

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

$W=F_d- F_f_r_id-F_gh$

$W=F_d-\mu_kmgdcos\theta-mgdsin\theta$

The change in kinetic energy is ,

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2$

At the top of the inclined plane , the velocity is zero

So,

$\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2$

$\Delta KE=-\frac{1}{2}m\nu_0^2$

From the work energy theorem , we have $W=-\Delta K$ in case of friction , so

$\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta$

$KE=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object A-

$KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object B

$KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta$

$KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)$

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

$KE_A >KE_B$

<u>Therefore , option A is correct .</u>

6 0

3 years ago

A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder

frozen [14]

<h2>Answer: $13.5\times 10^{8}$ joules</h2>

Explanation:

From the first law of thermodynamics,

Δ $Q$ =Δ $U$ + $W$

Where $Q$ is the heat given to the gas,

$U$ is the internal energy of the gas,

$W$ is the workdone by the gas.

When pressure is constant,

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}$

When pressure is constant, $W=P$ Δ $V$

Where $P$ is pressure and $V$ is the volume of the gas.

Given $P=1.5\times 10^{5}Pa$

Δ $V=$ $7.8-2.6=5.2m^{3}$

So, $W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J$

Given that Δ $U=6\times 10^{5}$

So,Δ $Q=$ $6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J$

5 0

3 years ago

The nutritional calorie (Calorie) is equivalent to 1 kcal. One pound of body fat is equivalent to about 4.1 x 103 Calories. If a

Luba_88 [7]

Answer:

$t = 70.13 hours$

Explanation:

As we know that 1 lb fat can be used to expend 4.1 k Calorie of energy

so here we know that we have to use 7.56 lb of body fat

so here the total energy that is to be utilized is given as

$Q = (4.1 \times 10^3)(7.56)$

so we will have

$Q = 30996 Calorie$

now we know that runner expends the energy at rate of 1850 kJ/h

so we have

$E = 1850 \times 10^3 kJ/h$

so it is

$E = 0.44 \times 10^3 calorie/h$

now we know that time to burn this energy is given as

$t = \frac{Q}{E}$

$t = \frac{30996}{0.44 \times 10^3}$

$t = 70.13 hours$

5 0

3 years ago

Plz help me I’m a desperate little boy

Neporo4naja [7]

It is the first one. But if you have this question you are not a little boy

5 0

3 years ago