Ask question

Ask question

All categories

2 years ago

9

Which animal have no teeth

2 answers:

Kamila [148]2 years ago

6 0

Anteaters they do not have teeth

dmitriy555 [2]2 years ago

5 0

Answer:

Ant eaters

Explanation:

Pls give me brainleist

You might be interested in

If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object

12345 [234]

Answer:

ΔK = 24 joules.

Explanation:

Δ $K =$ Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

$W = F$ * Δ $s$

Δ $s$ can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ $s$ = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

$W = < 3, 4 >$ * $< 4, 3 >$ = $(3*4)+(4*3)$ = $24$ J

3 0

11 months ago

Could a person at the south pole see the north star, explain??

9966 [12]

Answer:

If conditions are just right, you can see Polaris from just south of the equator. Although Polaris is also known as the North Star, it doesn't lie precisely above Earth's North Pole. If it did, Polaris would have a declination of exactly 90 degree.

Explanation:

4 0

2 years ago

Read 2 more answers

A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl

Dimas [21]

Answer:

<em>10.90km</em>

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

<em>d = 10.90km</em>

<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>

<em></em>

6 0

2 years ago

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface

Blizzard [7]

Answer:

$\rho = 7.35\times \muC/m^3$

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

$E = \dfrac{Rl}{2\epsilon_0}$

ε₀ is the permitivity of free space

R is the radius of the rod

and also,

$\rho=\dfrac{2E\epsilon_0}{R}$

ρ is the volume charge density

$\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}$

$\rho = 7.35\times 10^{-6}\ C/m^3$

$\rho = 7.35\times \muC/m^3$

Hence, the volume charge density is equal to $\rho = 7.35\times \muC/m^3$

6 0

2 years ago

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

So

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

3 years ago