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3 years ago

8

You want to calculate the displacement of an object thrown over a bridge. Using -10m/s^2 for acceleration due to gravity, what w

ould be the total displacement of the object if it took 8 seconds before hitting the water?

1 answer:

tekilochka [14]3 years ago

4 0

given that acceleration due to gravity is a = -10 m/s^2

initial speed will be ZERO

and we need to find the displacement in t = 8 s

now we can use kinematics equation to find the displacement

$\delta y = v_y * t + \frac{1}{2}at^2$

now plug in all values in it

$\delta y = 0* 8 + \frac{1}{2}*(-10)*8^2$

$\delta y = -320 m$

<em>so it will displace downwards by 320 m</em>

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The wasted energy in a charger is:

kirza4 [7]

Thermal is the wasted energy in a charger

4 0

3 years ago

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

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3 years ago

if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

Tresset [83]

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}$

where $\displaystyle{\vec{a}}$ is acceleration, $\displaystyle{\vec{v}}$ is velocity and $\displaystyle{t}$ is time. $\displaystyle{v_2}$ means final velocity. $\displaystyle{v_1}$ means initial velocity, $\displaystyle{t_2}$ means final time and $\displaystyle{t_1}$ means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

Our initial velocity starts at 0 mph.
Our final velocity is at 60 mph.
Our initial time is 0 second.
Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

A mile equals to 1609.344 meters.
An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

$\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}$

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

Initial velocity = 0 m/s
Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

$\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}$

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

3 0

1 year ago

A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang

valina [46]

Answer:

work done is -2.8 × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

$W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta$

$W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta$

$W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta$

W = $-mgl[$ -cosθ $]^{0.047}_{0.045 }$

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8 × 10⁻⁶ J

Therefore, work done is -2.8 × 10⁻⁶ J

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3 years ago

A block mass m (0.25 kg) is pressed against (but is not attached to) an ideal spring of force constant k (100 N/m) and negligibl

VMariaS [17]

Answer:

d. All the above choices are correct.

Explanation:

When a spring of spring constant k is compressed by distance x , the potential energy stored in it is equal to

E = 1/2 k x²

If spring constant is 2 k , potential energy stored

E = 1/2 2k x²

= k x²

which is twice the earlier potential energy.

In the first case , the energy of spring is imparted to box . The energy given to box is spent by frictional force due to which box comes to rest.

So energy of box acquired from spring = work done by frictional force.

So energy of box acquired from spring = F X d , F is frictional force , d is displacement .

In the second case ,

energy acquired by box becomes two times

Work done by frictional force will also become two times to put box at rest

So displacement will be two times ( because frictional force is constant )

so option a is correct .

option b is also correct .

Because kinetic energy of box will be twice as explained above .

So option d will be correct.

5 0

3 years ago