1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kondor19780726 [428]
2 years ago
13

An object is shot upwards, from the ground, with an initial velocity of 120

Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer: 30 metres

Explanation:

Initial velocity of object = 120m/s

Time taken = 4.0s

Distance covered by object = ?

Recall that distance = (Change in velocity / Time taken)

Distance = (120m/s)/4.0s

= (120m/s) / 4.0s

= 30m

Thus, the object will be 30 metres high

You might be interested in
5. How much does a 20 m x 10 m x 8 m swimming pool filled with water weigh? Assume that water has a density of 62 kg/m'.​
Nookie1986 [14]

Explanation:

First of all we calculate the volume which is 20m×10m×8m= 1600m^3 and then multiply the density by the volume to get the mass which will equal to 99200kg therefore the weight will be

<em><u>Weight</u></em>

mass × acceleration due to gravity

=99200kg × 10m^s^2

=992000N

7 0
2 years ago
Please answer these questions <br> 30 points
Maksim231197 [3]

Answer:

if i were you i would try to do the work because if you let someone else you wont be able to understand the question

6 0
3 years ago
Read 2 more answers
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

5 0
3 years ago
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1
ahrayia [7]

Answer:

The temperature is  T  = 168.44 \ K

Explanation:

From the question ewe are told that

   The rate of heat transferred is    P  = 13.1 \ W

     The surface area is  A = 1.55 \ m^2

      The emissivity of its surface is  e = 0.287

Generally, the rate of heat transfer is mathematically represented as

           H  =  A e \sigma  T^{4}

=>         T  =  \sqrt[4]{\frac{P}{e* \sigma } }

where  \sigma is the Boltzmann constant with value  \sigma  = 5.67*10^{-8} \ W\cdot  m^{-2} \cdot  K^{-4}.

substituting value  

             T  =  \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }

            T  = 168.44 \ K

7 0
3 years ago
Other questions:
  • If two sets of data are correlated, this means that:
    7·2 answers
  • Which type of energy is thermal energy a form of?
    15·1 answer
  • How does being underwater change the ability to localize sound? why is this so?
    6·1 answer
  • Name the physical quantity measured by the velocity-time graph?
    6·1 answer
  • Patients with brain tumors may elect to have a procedure known as Gamma Knife. In a typical Gamma Knife procedure gamma rays are
    9·2 answers
  • For each of the
    14·1 answer
  • What physics concept is used when designing a motorcycle helmet to protect the head from injury
    15·1 answer
  • An object moving at a constant velocity will always have a what
    10·2 answers
  • A body is dropped from a height of 30m. What is the velocity of the body after it has covered a distance of 20 m? (Given g= 10 m
    6·2 answers
  • Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!