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cestrela7 [59]
2 years ago
14

The graph shows the progress of a car which traveled 300 km in 5 hrs. Assuming that its speed remained constant, how far had the

car gone after 2 hrs?
Physics
1 answer:
lisov135 [29]2 years ago
6 0
If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km
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Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture
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Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

v_{0y}: is the initial vertical velocity of the net = 0 (it is dropped from rest)

0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}

t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

v = \frac{x}{t}

x = v*t = 18 m/s*1.75 m = 31.5 m

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!                                                                    

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2 years ago
Solve the science problem
scZoUnD [109]

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What is the problem I cant help unless you have the problem.

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If ?h°rxn and ?s°rxn are both negative values, what drives the spontaneous reaction and in what direction at standard conditions
irakobra [83]

The release of free energy drives the spontaneous reaction.

Spontaneity can be <span>determined using the change in </span>Gibbs free energy (the thermodynamic potencial):

delta G=delta H – T*delta S

where delta H is the enthalpy and delta S is the entropy.

The direction (the sign) of delta G depends of the changes of enthalpy and entropy. If delta G is negative then the process is spontaneous.

In our case, both delta H and delta S are negative values, the process as said is spontaneous which means that it may proceed in the forward direction.

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3 years ago
What is the momentum of an object that is moving in a straight line at a speed of 10 m/s and has a mass of 10 kilograms?
Aleksandr-060686 [28]
The momentum of an object is given by the product between its mass and its velocity:
p=mv
where m is the mass and v the velocity.

For the object in our problem, m=10 kg and v=10 m/s, therefore its momentum is
p=mv=(10 kg)(10 m/s)=100 kg m/s
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8 0
3 years ago
You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend
Bogdan [553]

Answer:

a.\tau=200J b.\alpha=0.44 \frac{rad}{s^2} c. \alpha=0.33\frac{rad}{s^2} d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by \tau=rF\sin \theta, being r the radius, F the force aplied and \theta the angle between the vector force and the vector radius. Since \theta=90^{\circ}, \, \sin\theta=1 and so \tau=rF=2m100N=200Nm=200J.

b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}.

c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0
3 years ago
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