Ask question

Ask question

All categories

vladimir2022 [97]

3 years ago

15

A couch is pushed with a force of 82 N and moves a distance of 6 m across the floor. How much work was done in moving the couch?

Round to a whole number.

1 answer:

erma4kov [3.2K]3 years ago

7 0

Answer:

492J

Explanation:

You might be interested in

PLEASE HELP I HAVE FINALSSS

vitfil [10]

Answer: 26.67 m/s

Explanation:

Given

Length traveled by the ball $s=40\ m$

Time taken to reach the goal post is $t=3\ s$

Initial velocity $u=0\ m/s$

Using the second equation of motion

$\Rightarrow s=ut+\frac{1}{2}at^2\\\Rightarrow 40=0+\frac{1}{2}a(3)^2\\\Rightarrow a=\frac{80}{9}\ m/s^2\\$

Now using

$\Rightarrow v^2-u^2=2as\\\\\Rightarrow v^2-0=2\times \frac{80}{9}\times 40\\\\\Rightarrow v=\sqrt{\dfrac{80\times 80}{9}}=\dfrac{80}{3}\\\Rightarrow v=26.67\ m/s$

The velocity of ball will be 26.67 m/s

8 0

3 years ago

A hockey puck is hit with a large force. The puck moves at a high speed as it slides over the ice.

ValentinkaMS [17]

According to Newton's first law of motion, the puck gradually slows down because friction would be acting on it.

<h3>What is Newton's first law?</h3>

Newton's first law of motion states that an object will remain at rest or move uniformly in a straight line unless acted upon by an external force.

According to this question, a hockey puck is hit with a large force. The puck moves at a high speed as it slides over the ice.

However, in accordance with the first law of motion proposed by Newton, the puck gradually slows down because friction would be acting on it.

Learn more about Newton's law at: brainly.com/question/974124

#SPJ1

8 0

2 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

True or False: In a p-n junction diode, the absolute value of the electric field is the largest at the metallurgical junction (e

Delvig [45]

Answer:

false

Explanation:

6 0

3 years ago

Why is training important?

stira [4]

So you work hard and you do well i’m getting good grades or doing something better.

7 0

3 years ago