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1 year ago

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

Physics

2 answers:

Alekssandra [29.7K]1 year ago

8 0

Answer:

2 meters per second²

Explanation:

PilotLPTM [1.2K]1 year ago

7 0

Answer:2m/s^2

Explanation:

mass=0.25kg

Force =0.5N

Acceleration=force/ mass

Acceleration=0.5/0.25

Acceleration=2

Acceleration =2m/s^2

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A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.

aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

$m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c}$ Where $m_{c}$ is the calorimeter mass and $n_{s}$ is the number of moles of the samples; $H_{c}$ is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product $m_{c} *Cp$ is the heat capacity, so the balance equation is:

$6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}$

So, the enthalpy of combustion can be calculated:

$H_{c}=-5899.5\frac{kJ}{mol}$

I will be happy to solve any doubt you have.

4 0

2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

8 0

1 year ago

A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0

Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

$vsin(\alpha) + gt = 0$

$t = \frac{-vsin(\alpha)}{g}$

So the vertical distance it travels within time t is

$y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}$

$y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}$

$y = \frac{-v^2sin^2(\alpha)}{2g}$

Similarly the horizontal distance it travels within time t is:

$x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}$

$x = \frac{-v^2sin(2\alpha)}{2g}$

We can pre-calcualte the quantity $\frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54$

So $y = 57.54sin^2(\alpha)$

$x = 57.54sin(2\alpha)$

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0

1 year ago

A parcel is traveling on a horizontal conveyer belt moving at 1 meter/second. At the end of the conveyor belt, the parcel lands

Igoryamba

Assume the acceleration due to gravity is approximately 9.8 m/s^2

In 1 second, it will fall

s = ut + at^2/2

u = initial vertical velocity = 0

a = 9.8

t = 1

s is the distance it falls

= 9.8/2 = 4.9 m

But the parcel has a horizontal velocity of 1 m/s so it will travel 1 m horizontally in 1 second.

The answer would be

1 metre, 4.9 metres.

6 0

2 years ago

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of

son4ous [18]

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0

1 year ago