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serious [3.7K]
3 years ago
10

If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft

er t seconds is given by the equation h equals negative 16 t squared plus 112 t plus 128. After how many seconds does the object hit the​ ground? Round to the nearest tenth of a second.
Physics
1 answer:
Marina CMI [18]3 years ago
8 0

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

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Flasher units are being discussed. Technician A says that only a DOT-approved flasher unit should be used for turn signals. Tech
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. A student thinks that the more protons an atom has, the more reactive it will be. Which will best disprove this student's idea
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3 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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3 years ago
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Answer:

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