Where was Malala attacked Where was

POR FAVOR AYUDENME A COMLETARLO

Zigmanuir [339]

Answer:

English please?

Explanation:

8 0

3 years ago

How to find out the heat capacity of a material?

DochEvi [55]

$\huge\underline{\underline{\boxed{\mathbb {EXPLANATION}}}}$

The heat capacity is given by the expression:

$\longrightarrow \sf{\triangle Q= m \triangle C \triangle T}$

$\longrightarrow \sf{Q= \: Heat}$

$\longrightarrow \sf{M= \: Mass}$

$\longrightarrow \sf{C= \: Specific \: Heat}$

$\longrightarrow \sf{T= \: Temperature}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\leadsto$ When the $\bm{heat}$ is measured in the calorimeter, we obtain a value, and since we know the mass of the material and we control the change in $\bm{temperature}$ , we can then determine the specific heat "C" by simply remplazing in the expression.

5 0

2 years ago

The physical structure of the earth’s rock is changed by _____.

PSYCHO15rus [73]

It could be stress or strain

4 0

4 years ago

Read 2 more answers

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

$A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\$

since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

6 0

4 years ago

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45

zheka24 [161]

Answer:

v = 12 m/s

Long, boring, and convoluted explanation:

First, let's lay out our information.

- max height = 3.7 m

- 0 = 45°

- gravitational acceleration constant = 9.8 $\frac{m}{s^2}$

Since the puma leaves the ground at a 45 ° angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is 3.7m because the puma can reach up to that height. Gravity is always given to be 9.8 $\frac{m}{s^2}$

Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity, vsin ( θ ) , and not the horizontal component, vcos ( θ ) .

Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: vsin ( θ ) = $\sqrt{2hg}$

Where vsin ( θ ) is the vertical component of the initial velocity and h and g are max height and gravitational acceleration constant respectively.

Plugin, rearrange and solve

v sin ( θ ) = $\sqrt{2hg}$

v sin ( 45 ∘ ) = √ 2 × 3.7 × 9.8

v ( 0.71 ) = $\sqrt{72.52}$

v ( 0.71 ) = 8.52

v = 8.52 /0.71

v = 12 m s

4 0

3 years ago