Your’re working with a patient who suddenly falls. You should

A runner starts at point A, runs around a 1-mile track and finishes the run back at point A. Which of the following statements i

Lyrx [107]

This question is incomplete because the options are missing; here is the complete question:

A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?

A. The runner's displacement is 1 mile.

B. The runner's displacement is zero.

C. The distance the runner covered is zero.

D. The runner's speed was zero.

The answer to this question is B. The runner's displacement is zero

Explanation:

Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.

4 0

3 years ago

PLEASE HELP<br> 8th grade honors science

lora16 [44]

Both of them are magnets coiled by wires.

1. The wire coiled in the first diagram, the wire is having current, Making the magnetic feild of the magnet more........

2. The wire coiling the magnet is here not having electric current making the magnetic feild smaller

7 0

3 years ago

Read 2 more answers

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0

3 years ago

A jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in mete

lorasvet [3.4K]

Answer:

The acceleration is $a =2\ m/s^2$

Explanation:

From the question we are told that

The mass of the jumbo jet is $m_j = 100000\ kg$

The thrust is $F_k = 50000 \ N$

Generally given that the jet has four engines the total thrust is

$F_t = 4 * F_k$

substituting values

$F_t = 4 * 50000$

$F_t = 200000 \ N$

Generally the acceleration of the is mathematically represented as

$a = \frac{F_t}{m}$

substituting values

$a =2 \frac{N}{kg}$

Now

$N = kg \cdot m/s^2$

Hence

$a =2 \frac{kg * \cdot m/s^2}{kg}$

$a =2\ m/s^2$

5 0

3 years ago

For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What

choli [55]

Answer

given,

Stress for plastic deformation = 267 MPa

modulus of elasticity = 115 GPa

cross sectional area = 377 mm²

a) maximum load (in N) that may be applied to a specimen

= σ x A

= 267 x 10⁶ x 377 x 10⁻⁶

= 100659 N

b) modulus of elasticity = stress/strain

115 x 10⁹ = $\dfrac{267 \times 10^6}{\dfrac{\Delta l}{L}}$

L = 127 mm

115 x 10⁹ = $\dfrac{267 \times 10^6}{\dfrac{\Delta l}{127}}$

$\dfrac{\Delta l}{127}=\dfrac{267 \times 10^6}{115\times 10^9}$

Δ l = 0.295 mm

maximum length after the stretched = 127 mm + 0.295 mm

= 127.295 mm

4 0

3 years ago