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ehidna [41]
3 years ago
11

Yasmin is testing an unknown solution to determine whether it is an acid or a base. She dips red litmus paper in the solution. I

t remains red. What can she conclude about the sample?
Physics
2 answers:
victus00 [196]3 years ago
7 0
The correct answer is D) It is not a base
nydimaria [60]3 years ago
5 0

the correct answer is d


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A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si
Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B)         -  Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m;  c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b  = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0
3 years ago
uppose two train cars are moving toward one another, the first with a mass of 150,000 kg and a velocity of 0.300 m/s; the second
kondaur [170]

The value was determined to be 0.122 m/s. The velocity of a body or object determines its direction of motion. Speed is a scalar quantity in its most fundamental form.

Velocity is essentially a vector quantity. It is the rate of change in distance. The initial speed of the first train, which has a mass of 150,000 kg, is 0.3 m/s. The second train has an initial speed of -0.120 m/s and a mass of 110,000 kg.

Let v represent the post-collision speed of the connected mass.

Utilize the idea of momentum.

The speed of the trains is constant both before and after a collision.

150.000 + 110.000v 45.000 - 13200 = 260.000 v 31800 = 260.000 v v = 0.122 m/s 150000 x 0.3 - 110000 x 0.120

After colliding, they move at a speed of 0.122 m/s towards the direction of the right.

Learn more about velocity here-

brainly.com/question/18084516

#SPJ4

7 0
1 year ago
How many electrons can exist on the following energy levels
alukav5142 [94]
2e level 1
8e level 2
18e level 3
32e level 4
6 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
As seen in the figure, a bullet with mass of 15.0-g is fired vertically and penetrates a block with mass of 2.5-kg and the block
rodikova [14]

Answer:

KE = 2.03 J

Explanation:

After impact, the kinetic energy of the bullet+block will convert to potential energy

½mv² = mgh

v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s

conservation of momentum during the collision

0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)

u = 16.4481 m/s

KE = ½mv² = ½(0.015)16.4481² = 2.0290499...

KE = 2.03 J

4 0
3 years ago
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