Physics, 15.12.2019 12:31 SophomoreSareke

In a football game, one team kicks off to the other. at the moment the receiver catches the ball, he is 40 yards from the nearest tackler. the receiver runs left to right at a speed of 10 yards per second (10 y/s). the tackler runs right to left at a speed of 6 yards per second.

how long does it take before they collide?

how far does the receiver go?

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Answer from: cavejr87

The time the two players will collide is 2.5 s

The distance traveled by the receiver before the collision is 25 yards.

The given parameters;

distance the two players, = 40 yardsvelocity of the receiver, Vc = 10 yards/svelocity of the tackler, Vt = 6 yards/s

To find:

the time before they collidethe distance traveled by the receiver

The time taken before the two players collide is calculated by applying relative velocity principle;

Let the time when they collide = tassume left to right = positiveright to left = negative

$(V_c - (-V_t))\times t = 40 \ yards$

$(10 + 6)t = 40\\\\16t = 40\\\\t = \frac{40}{16} \\\\t = 2.5 \ seconds$

The time the two players will collide is 2.5 s

The distance traveled by the receiver before the collision is calculated as;

$distance = speed \times time\\\\distance = V_c \times t\\\\distance = 10 \times 2.5= 25 \ yards$

Thus, the distance traveled by the receiver before the collision is 25 yards.

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Answer from: jjjjj23407

Part a)

$\delta t = 2.5 s$

Part b)

d = 25 yards

Explanation:

Speed of the receiver is 10 yards per second towards right

Speed of the tackler is 6 yards per second towards left

So here we have net relative speed of the two towards each other is given as

$v_{12} = v_1 - v_2$

$v_{12} = 10 - (-6) = 16 yards/s$

now we know that the distance between receiver and tackler initially is 40 Yards

Part a)

so the time taken by two to collide is given as

$\delta t = \frac{\Delta x}{v}$

$\delta t = \frac{40}{16}$

$\delta t = 2.5 s$

Part b)

Distance moved by the receiver in t = 2.5 s

d = vt

d = 10 (2.5)

d = 25 yards

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

the answer is a. at the equator

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