Someone please help me!

A dog in an open field runs 11.0 m east and then 28.0 m in a direction 49.0 ∘ west of north. In what direction must the dog then

Ede4ka [16]

Answer:

$\theta=19.03 SE$

Explanation:

From the question we are told that:

$d_1=11.0m East$

$d_2=28.0m 49 \textdegree west\ north$

$d_3=10m south$

Generally the equation for Resolutions is mathematically given by

$d_x=-11-(-28sin49)$

$d_x=10.132m$

Where

$d_y=-11-(28cos49)$

$d_y=-29.37m$

Generally the equation for Direction is mathematically given by

$\theta=tan^{-1}(\frac{dx}{dy})$

$\theta=tan^{-1}(\frac{10.132}{29.37})$

$\theta=19.03 SE$

7 0

3 years ago

Which of the following is a list of

KonstantinChe [14]

Answer:

c. metallic, sub metallic, or nonmetallic

4 0

3 years ago

A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After

Gala2k [10]

Answer with Explanation:

We are given that

Mass of bullet, $m_1=0.03 kg$

$u_1=470 m/s$

$m_2=3 kg$

$\mu_k=0.2$

$m_3=30 kg$

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

$m_1u_1=(m_1+m_2)v$

$0.03(470)=(0.03+3)v$

$v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s$

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

$(m_1+m_2)v=(m_1+m_2+m_3)V$

$(0.03+3)\times 4.65=(0.03+3+30)V$

$V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}$

$V=0.43 m/s$

3 0

4 years ago

Read 2 more answers

Would a rougher or smoother object absorb thermal energy faster? Why? Give an example.

Nuetrik [128]

Smoother because it will increase energy and when the energy increases it’ll create heat also . Example: A car racing on a smooth road it’ll go faster than a Car speeding on a bumpy and rough road , Hope that helps .

5 0

4 years ago

A cyclist makes the following trip along two vectors; he travels 9km to the north and then travels 6km to the east

elena-14-01-66 [18.8K]

Answer:

Final distance from the origin: 10.82 km. the vector points as shown in the attached image.

Angle with respect to the east: $56.31^o$

Explanation:

Please refer to the attached image. The cyclist's trip is indicated with the green arrows (9 km to the north followed by 6 km to the east.

So his final position is at the tip of this last vector, and indicated by the orange vector drawn form the point where the trip starts to the cyclist's final location.

We observe that this orange vector is in fact the hypotenuse of a right angle triangle, and we can estimate the distance from the origin by the Pythagorean theorem:

$d=\sqrt{9^2+6^2} \\d=\sqrt{81+36} \\d=\sqrt{117} \\d=10.82 \,\,km$

Notice that this is NOT the actual number of km that the cyclist pedaled to reach the final point.

Now, to find the value of the angle $\theta$ , we need to use trigonometry, and in particular the tangent function gives us the ratio between the side of the triangle "opposite" to the angle, divided the side "adjacent" to the angle:

$tan(\theta)=\frac{opp}{adj} \\tan(\theta)=\frac{9}{6}\\tan(\theta)=\frac{3}{2}\\$

Now we can find the value of the angle by using the arctan function:

$tan(\theta)=\frac{3}{2} \\\theta=arctan(\frac{3}{2} )\\\theta= 56.31^o$

6 0

3 years ago