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1 year ago

15

WHAT IS THE DIRECT PATH DIRECTION FROM POINT A TO B

2 answers:

nasty-shy [4]1 year ago

7 0

Answer:i want to say route but there is not enough information

Explanation:

PolarNik [594]1 year ago

5 0

I would gladly help you,but I would need more information to help

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod

Scorpion4ik [409]

Answer: $-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$2Br^.\rightarrow Br_2$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{d[Br^.]}{2dt}$

or $Rate=+\frac{d[Br_2]}{dt}$

Thus $-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}$

4 0

3 years ago

In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.

JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0

3 years ago

You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in

Vlad1618 [11]

Answer:

$0.28 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

$u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s$ is the initial velocity

$v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s$ is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

$a=\frac{2.78-2.22}{2}=0.28 m/s^2$

5 0

4 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Vlad [161]

Answer: $7.95^{\circ}C$

Explanation:

Given

Mass of water is $m_1=1\ kg$

mass of ice is $m_2=0.1\ kg$

Latent heat of fusion $L=3.33\times 10^5\ kJ/kg$

The heat capacity of water is $c_{w}=4186\ J/kg^{\circ}C$

Suppose water is at $T^{\circ} C$ and it reaches to $0^{\circ}C$ to melt the ice

the heat released by water must be equivalent to heat absorbed by the ice

$\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C$

6 0

3 years ago