Please answer fast ———

______ had occurred when your body can no longer regulate its temperature

sergejj [24]

Heat stroke had occurred when your body can no longer regulate its temperature. hope it helps :)

4 0

3 years ago

Timmy bikes to school 3 kilometers east and 2 kilometers south. What is Timmy's final displacement?

vovikov84 [41]

Answer:

3.6km South East

Explanation:

Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:

Pythagoras Theorem:

a^2+b^2=c^2

(2)^2+(3)^2=c^(2)

4+9=c^2

Square root 13 = c

c=3.6km (1dp)

The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).

Hope this helped!

8 0

3 years ago

An iron anchor of density 7810.00 kg/m^3 appears 252 N lighter in water than in air. (a) What is the volume of the anchor?(b) Ho

RSB [31]

Answer:

(a)The volume of the iron anchor = 25.2 m³

(b)The mass of iron = 196812 kg

Explanation:

Density: This is the ratio of the mass of a body to its volume. The S.I unit of density is kg/m³

(a)

From Archimedes principle, the iron will displace a volume of water that is equal to its volume. And the weight of water displaced is equal to loss in weight or upthrust.

Therefore, Upthrust = lost in weight = weight of water displace.

Therefore, Volume of the iron = mass of water displace/density of water

Vi = m₁/D₁........................... Equation 1

Where Vi = volume of iron, m₁ = mass of water displaced, D₁ = Density of water.

Given: upthrust = 252 N, ∴ m₁ = 252/10 = 25.2 kg.

Constant: D₁ = 1.0 kg/m³

Substituting these values into equation 1,

Vi = 25.2/1.0

Vi = 25.2 m³

Therefore the volume of the iron anchor = 25.2 m³

(b)

Density of the iron = mass of the iron/volume of the iron.

Therefore, Mass of the iron = Density of the iron × volume of the iron...........................................equation 2

Given: Density of the iron = 7810 kg/m³, Volume of the iron = 25.2 m³

Substituting these values into equation 2,

Mass of the iron = 7810×25.2

Mass of the iron = 196812 kg.

Therefore the mass of iron = 196812 kg

8 0

3 years ago

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Naya [18.7K]

Answer:

$2.966\times 10^{-11}\ N$

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = $6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}$

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

$F=\dfrac{GMm}{d^2}$

Plug in the given values and solve for 'F'. This gives,

$F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})$

The same force is experienced by both cannonball and marble.

Therefore, the gravitational force of the marble is $2.966\times 10^{-11}\ N$

3 0

3 years ago

A monochromatic light of wavelength 506 nm from a distant source is incident on a single slit 0.075 mm wide. Diffraction pattern

sladkih [1.3K]

Answer:

first minimum distance from the center = 0.417 m m

Explanation:

given,

wavelength (λ) = 506 nm

d = 0.075 mm = 0.000075 m

tanθ = x /2

d sinθ = m λ

m=1 ( 1 st minimum )

0.000075 × sin θ = 506 × 10⁻⁹

sin θ = $\dfrac{506 × 10⁻⁹}{0.000075}$

sin θ = 0.00675

θ = sin⁻¹(0.00675)

θ = 0.00675

x = 0.000417 m

hence, the first minimum distance from the center = 0.417 m m

5 0

3 years ago

Please answer fast ————-