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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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Two cars, one of mass 1400 kg, and the
Vikki [24]

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a   ⇒   a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b   ⇒   b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

3 0
2 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
3 years ago
Three boys of equal strength try to break a rope (and fail) by tying one end to a fence post and tugging on the other end. Three
podryga [215]

The best choice would be C. Keep one end of the rope tied to the post and have all six boys tug on the other end.

Reason:

A is not correct since in the first case which is B, untying the rope from the post and have 3 boys on each end of it tug, will provide equal or a near equal amount of tension since the boys are all within the same strength, however there can be variables that will affect the tension.

That means B would be true according to A, and then for C, having all six boys pull on the other end of the rope would provide greater tension, since if 3 boys of the same strength range can't break the rope when its tied to the fence, we add the other 3 boys of the same strength.

However, if 3 boys are on each end of the rope and of the same strength then the rope will not break since there is an equal amount of net force exerted on the rope.

We know that the fence can withstand the strength of 3 boys but if we add the other 3 boys then it could provide us with a different outcome.

3B < F

3B(2) ≥ F

So, therefore, our best choice would be C since A and B is incorrect.

4 0
3 years ago
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PolarNik [594]

Answer:

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2 years ago
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8 0
3 years ago
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