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2 years ago

Classify the animal hydra and plant mustard with two characters of each

Physics

1 answer:

inna [77]2 years ago

4 0

Answer:

Mustard seeds, both white and brown, are nearly globular in shape, finely pitted, odourless when whole, and pungent-tasting

Image result for character of hydra

Introduction: Hydra are inconspicuous freshwater relatives of corals, sea anemones and jellyfish. All are members of the phylum Cnidaria, characterized by radially symmetrical bodies, presence of stinging tentacles and a simple gut with only one opening (gastrovascular cavity)

Explanation:

Hydra are a genus of small, fresh-water organisms that are classified under the phylum Cnidaria.the classification of mustard is

Dicotyledons

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How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

11 months ago

A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho

Lemur [1.5K]

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

6 0

2 years ago

vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di

pantera1 [17]

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

6 0

3 years ago

A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,

kirill115 [55]

Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

θ = 90°,

work done = F s cos 90° ∵ cos 90° = 0

Work done = 0 J

8 0

3 years ago

A party shop delivers helium-filled balloons to homes and businesses. The owners realize from experience that on hot summer days

kvasek [131]

Answer:As the temperature increases, the helium in the ballon expands.

Explanation:

I took the quiz already

6 0

2 years ago

Read 2 more answers

Classify the animal hydra and plant mustard with two characters of each​

Classify the animal hydra and plant mustard with two characters of each