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3 years ago

1 Verify the identity. Show your work. cot θ ∙ sec θ = csc θ

1 answer:

6 0

To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos θ
csc θ = 1 / sin θ

Using these identities:
cot θ ∙ sec θ = (cos θ / sin θ ) ( 1 / cos θ)

We can cancel out cos θ, leaving us with
cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc θ

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An object that has the ability to do work has __________ energy. (4 points)

SpyIntel [72]

Answer:

c. potential

Explanation:

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3 years ago

Read 2 more answers

Find the components to write this vector in unit vector notation: 63.5 A please help

IrinaVladis [17]

Vector is perpendicular to x axis or i component.

Hence i component is 0

j component is 63.5

$\\ \sf\longmapsto \overrightharpoon{A}=0\hat{i} +63.5\hat{j}$

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3 years ago

Critical angle of glass is 42 .what does it mean?

elena55 [62]

Answer:

i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!

Explanation:

HOPE IT HELPS!!

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3 years ago

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

x = ½ a₁ t²

x = ½ a₂ (t-4)²

Let's solve

a₁ t² = a₂ (t-4)²

a₁/a₂ t² = t² -2 4 t + 16

t² (1- 2.0 / 4.0) - 8 t +16

t² 0.5 - 8 t +16 = 0

t² -16 t + 32 = 0

Let's solve the second degree equation

t = [16 ±√( 16² - 4 32)] / 2

t = ½ (16 ± 11,3)

Solutions

t1 = 13.66 s

t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0

3 years ago

2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to

ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b) the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

F = m a

a = F / m

Let's use kinematics to find the velocity after x = 1 m

v² = v₀² + 2 a x

The initial speed is zero

v = √ (2 F/m x)

For the distance of x = 1 m

v₁ = √ (2 F / m)

Car 2

F = 2m a

a = F / 2m

v² = 2 a x

v = √ (F/m x)

For x = 1 m

v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m 2F / m

K₁ = F

Car 2

K₂ = ½ 2m v₂²

K₂ = ½ 2m F / m

K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m √ (2F / m)

Car 2

P₂ = 2m v²

P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

v = v₀ + a t

v = F / m t

v₁ = F / m 10

Car 2

v₂ = F / 2m 10

v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m (F / m 10)²

K₁ = 50 F² / m

Car2

K₂ = ½ 2m v₂²

K₂ = m (F / m 5)²

K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m F / m 10

P₁ = 10 F

Car 2

P₂ = 2m v₂

P₂ = 2m F / m 5

P₂ = 10 F

In this case the moment of the two cars is the same

7 0

3 years ago