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2 years ago

15

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop

. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What’s the acceleration of the wooden block when it hits the sensor? Use F = ma.

2 answers:

lilavasa [31]2 years ago

6 0

Answer:

a= 9.8

Explanation:

anastassius [24]2 years ago

4 0

To calculate the acceleration of the wooden block, we use the expression F=ma where F is the force applied, m is the mass of the object and a is the acceleration. We calculate as follows:

F = ma
4.9 = 0.5a
a = 9.8

Hope this answers the question. Have a nice day.

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

b)

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o

mariarad [96]

Answer:

$S_i=\frac{9}{10} =0.9$

Explanation:

Given:

volume of oil in the cylinder, $V_o=60\ cm^2$

volume of the oil level when the ice is immersed, $V=90\ cm^3$

the volume level of oil when the ice melted, $V'=87\ cm^3$

<u>Now, therefore the volume of ice:</u>

$V_i=V-V_o$

$V_i=90-60$

$V_i=30\ cm^3$

<u>Now the volume of water:</u>

$V_w=V'-V_o$

$V_w=87-60$

$V_w=27\ cm^3$

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

$S_i=\frac{\rho_i}{\rho_w}$ .....................(1)

as we know that density is given as:

$\rm \rho=\frac{mass}{volume}$

now eq. (1)

$S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}$

where, m = mass of the water or the ice which remains constant in any phase

$S_i=\frac{V_w}{V_i}$

$S_i=\frac{27}{30}$

$S_i=\frac{9}{10} =0.9$

7 0

2 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

sergejj [24]

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

3 0

2 years ago

Serena is a research student who has conducted an experiment on the discoloration of marble. Read about Serena’s experiment. The

sergij07 [2.7K]

The two flaws in her experiment’s design are

<span>- She introduced at least one confounding variable.</span> <span>- She tried to test multiple hypotheses at a time</span>

In the above mentioned experiment she had to have four samples to prove four hypotheses, each one separately and not to mix two hypotheses in an alone sample, that what it brings as consequence is the confusion.

3 0

2 years ago

A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height

Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

$v_x = 23.2 m/s$

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

$v_y(t)= v_{y0} -gt$ (1)

where

$v_{y0}=0$ is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

$h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s$

Substituting t into (1) we find the final vertical velocity

$v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s$

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

$tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}$

6 0

2 years ago