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Pie
3 years ago
13

Ted is a high school student. He is very good at math, but he is not a very good English student. He likes all of his teachers;

he thinks they are fair and want him to learn. On Tuesday, Ted was sitting in his English classroom, listening to his teacher’s lecture. The teacher asked the class a question. He called on Ted. Ted had not been paying attention. His palms began to sweat, and his heart rate increased. He felt embarrassed that he did not know the answer. On Wednesday, Ted was asked to answer another question in his math classroom. Even though this time he was listening and knew the answer, his palms began to sweat and his heart rate increased. Why did Ted’s palms sweat and heart rate increase on Wednesday? A. Ted associated his embarrassment with being in high school. B. Ted associated his math teacher with feelings of embarrassment. C. Ted was not completely sure that his answer was correct. D. Ted associated being asked a question with embarrassment.
Physics
2 answers:
oee [108]3 years ago
7 0

Answer:

I think its D...?

Explanation:

TAking the test rn

scoray [572]3 years ago
3 0
I think the correct answer is

D) Ted associated being asked a question with embarrassment.

Glad I could help, and good luck!

AnonymousGiantsFan
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What units would you use to measure the amount of liquid inside a can of soda?
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3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine
DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

F_x = F cos \theta

with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

F_f = \mu mg

where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N

8 0
3 years ago
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