All categories

3 years ago

6. A physics book slides off a horizontal table top with a speed of 1.25m/s. It strikes a floor in 0.4s.

Physics

1 answer:

aliina [53]3 years ago

8 0

Answer:

4.11 m/s

Explanation:

Given:

v₀ₓ = 1.25 m/s

aₓ = 0 m/s²

v₀ᵧ = 0 m/s

aᵧ = -9.8 m/s²

t = 0.4 s

Find: v

First, in the x direction:

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.4 s) + 1.25 m/s

vₓ = 1.25 m/s

And in the y direction:

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-9.8 m/s²) (0.4 s) + 0 m/s

vᵧ = -3.92 m/s

The net speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1.25 m/s)² + (-3.92 m/s)²

v = 4.11 m/s

You might be interested in

In this case, lithium would be classified as a(n)

Black_prince [1.1K]

Lithium is to be considered as an ION

Atom is defined as its natural state where it will have all its electrons present in its shell

Since in this case Lithium is giving its electron so after giving the electron it will become an ion

Compound is the combination of two or more different atoms

Molecules is the combination of same type of atoms

So here answer must be

<em>C) ion</em>

5 0

4 years ago

Read 2 more answers

Plants need to....................synthesize

koban [17]

Answer:

nitrogen

Explanation:

because I also had this in exam and I was correct

5 0

3 years ago

By what factor will the acceleration of an object change its mass stays constant but the force acting on it triples

Viktor [21]

Answer:

lacceleration is given by

a=F/m

With force acting on the object remained constant ,the acceleration is inversly proportional to mass of the object i.e with increase in mass =>acceleration decreases and with decrease in mass => acceleration increases.

acceleration decreases

acceleration increase

7 0

3 years ago

Who else has 3 20s hmmmmmmm

hichkok12 [17]

dollar bills if so def. not me but grades 4

3 0

3 years ago

Read 2 more answers

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