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3 years ago

13

Please help with this!

1 answer:

Goshia [24]3 years ago

8 0

Answer:

The balloons have the same charge,

If they had opposite charges they would cling to one another.

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A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

8 0

4 years ago

Below is a physics question

Allisa [31]

Explanation:

i1=1ampere

i1+i2=I

VA−VB=i1×3Ω=i2×6

⇒1×3=i2×6

2i=21ampere

Hence I=i1+i2=1+0.5=1.5ampere

Req=2+3+63×6=2+93×6=4Ω

equivalent circuit

Refer image .2

From KVL

1req=v

v=1.5×4

8 0

3 years ago

20 kg who is running at a speed of 4.0 m/s jumps onto a stationary sled of mass 5.0 kg on a frozen lake. the speed at which the

Step2247 [10]

Here we can use momentum conservation as there is no external force on sled and child while he jump on sled

by momentum conservation equation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v{2f}$

$20*4 + 5*0 = 20*v + 5 *v$

since sled and child both moves with same speed so here they both will have same final speed "v"

by solving above equation we will have

25 v = 80

v = 3.2 m/s

So they will move together with speed 3.2 m/s

6 0

3 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Solar energy can be changed into thermal energy without any work being done. true or false

Black_prince [1.1K]

Answer:

True

Explanation:

3 0

3 years ago