All categories

3 years ago

HELLLPPP FAST PLS

Physics

1 answer:

r-ruslan [8.4K]3 years ago

3 0

Answer:

Oh I had this question before! :)

Explanation:

Acid - citrus products - the classic one is lemon juice

Anything with a sour taste is acidic

Base - soap, oven cleaner, drain cleaner, baking soda

Basic things are bitter (please don't taste the drain cleaner and oven cleaner)

You might be interested in

A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T

ValentinkaMS [17]

Answer:

7.344 s

Explanation:

A = 0.15 x 0.3 m^2 = 0.045 m^2

N = 240

e = - 2.5 v

B1 = 0.1 T

B2 = 1.8 T

ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T

Δt = ?

e = - dФ/dt

e = - N x A x ΔB/Δt

- 2.5 = - 240 x 0.045 x 1.7 / Δt

2.5 = 18.36 / Δt

Δt = 7.344 s

6 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

What is the acceleration of the object?<br> m/s²

Orlov [11]

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

7 0

2 years ago

If a vehicle accelerating at 2.7 m/s2what is it's velocity at 20 meters 12.63m/s,1.03m/s,10.39m/s,6.39m/s

netineya [11]

Answer:

The final velocity of the vehicle is 10.39 m/s.

Explanation:

Given;

acceleration of the vehicle, a = 2.7 m/s²

distance moved by the vehicle, d = 20 m

The final velocity of the vehicle is calculated using the following kinematic equation;

v² = u² + 2ah

v² = 0 + 2 x 2.7 x 20

v² = 108

v = √108

v = 10.39 m/s

Therefore, the final velocity of the vehicle is 10.39 m/s.

5 0

3 years ago

When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end

Sunny_sXe [5.5K]

1) 29.8 C

At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).

2) 6.2 C

The temperature change of the water is given by the difference between its final temperature and its initial temperature:

$\Delta T = T_f - T_i$