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Nonamiya [84]
3 years ago
7

Diana and Kinsey are put in charge of choosing a mascot for their basketball team. There are fifteen players on the team, but Di

ana and Kinsey survey only the five players on the starting lineup about their favorite mascot. Which best describes this error in data collection?\
Physics
2 answers:
Vikentia [17]3 years ago
8 0

Answer:

  • The data are inadequate

Explanation:

rusak2 [61]3 years ago
6 0
<span>The data are inadequate.</span>
You might be interested in
What can one say for sure about element #99, Einsteinium (Es))? a) There's no such element! b) It is an alkali metal. c) It is r
Dmitrij [34]

Answer:

Option (C)

Explanation:

Einsteinium is an element of the periodic table grouped in the Actinide series, with atomic number 99. They are dense element and highly electro-positive. <u>They are highly radioactive</u>, i.e the atoms within the element are unstable and constantly decay until they reach a stable environment. It has 99 number of electrons and protons, 153 number of neutrons.

Due to its high radioactivity, they are health hazardous and can used in making nuclear weapons but their uses are very limited and unknown.

Thus, the correct answer is option (C).

8 0
3 years ago
How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h
tresset_1 [31]

Answer:

The  time taken is  t =  32.5 \  s

Explanation:

From the question we are told that

   The  speed  of  first car is  v_1  =  66.7 \ km/h  =  18.3 \  m/s

    The  speed  of  second car is v_2  =  52.7 \ km/h  =  14.64 \  m/s

   The  initial distance of separation is  d =  119 \ m

The distance covered by first car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  0 m/s

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_1 * t

So

     d_t =  0 \  m/s  +  (v_1 * t )

     d_t =  0 \  m/s  +  (18.3 * t )

The distance covered by second  car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  119 m

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_2* t

       d_t =  119  + 14.64 *  t

Given that the two car are now in the same position we have that

    119  + 14.64 *  t  =   0   +  (18.3 * t )

   t =  32.5 \  s

6 0
3 years ago
A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont
Georgia [21]

Answer:

141.30grams

Explanation:

First, denote our known values;

J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?

Mass is impulse divided by change in velocity:

m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs

Hence, the mass of the ball is 141.30grams

3 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

            Force =  Spring constant x Elongation

            F = kx

Here force is weight of mass

           F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 0
3 years ago
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