A liquid thermometer can be used to test for fevers. When body temperature increases,

One way to improve food security is to avoid food spoilage. Identify the correct ways that Devon reduced food spoilage in the pa

quester [9]

The correct answer would be "He brought one serving to his neighbor's house, and stored the other two servings in the refrigerator. Devon ate one more serving or spaghetti the following day."

3 0

3 years ago

Read 2 more answers

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

Pour chacune des réactions chimiques

Doss [256]

Answer: Me non espanol

Explanation:

4 0

2 years ago

The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the

8090 [49]

Compute first for the vertical motion, the formula is:

y = gt²/2

0.810 m = (9.81 m/s²)(t)²/2

t = 0.4064 s

whereas the horizontal motion is computed by:

x = (vx)t

4.65 m = (vx)(0.4064 s)

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed.

(vy) = gt

(vy) = (9.81 m/s²)(0.4064 s)

(vy) = 3.99 m/s

speed with which it hit the ground:

v = sqrt[(vx)² + (vy)²]

v = sqrt[(11.44 m/s)² + (3.99 m/s)²]

v = 12.12 m / s

6 0

3 years ago

A 5kg ball is on top of the school building at a height of 40m above the ground.

mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0

2 years ago