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1 year ago

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A professor determined the relationship between the time spent studying (in hours) and performance on an exam.

1 answer:

lubasha [3.4K]1 year ago

3 0

Lets write again formula for determening Ann's performance.

P = 70.443 + 4,885*t

where t is in hours. This is equation with P(t) which means that P only depends on variable t. If we express t=2.6 in formula we will find her expected performance.

P = 83.144

Now, since it says that she scored 16 points less than expected we need to find value of P-16

P - 16 = 67.144

After round we get that the answer is 67

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Read 2 more answers

Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration

worty [1.4K]

Answer:

3433.5 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of person = 70 kg

According to the question

a = Acceleration

$4g=4\times 9.81\\\Rightarrow a=39.24\ m/s^2$

Balancing the forces we have

$F-w=ma\\\Rightarrow F=ma+w\\\Rightarrow F=ma+mg\\\Rightarrow F=m(a+g)\\\Rightarrow F=70(39.24+9.81)\\\Rightarrow F=3433.5\ N$

The required force is 3433.5 N

3 0

1 year ago

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

azamat

Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

6 0

2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

3 0

1 year ago