All categories

2 years ago

Iron and steel went out c

Physics

1 answer:

Anton [14]2 years ago

7 0

Explanation:

a) 50km

b) 6hrs

c) speed a to b is 10km/h

b to c is rest

c to d is 20km/h, hence fastest.

d) 2 hours

You might be interested in

A pilot heads his jet due east. The jet has a speed of 425 mi/h relative to the air. The wind is blowing due north with a speed

Alja [10]

Answer:

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is, ∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

Explanation:

We can represent east direction by i and north direction by j.

The jet has a relative speed of 425 mi/h relative to the air.

The wind is blowing due north with a speed of 35 mi/h = 35 j

425 mi/h is the relative speed with respect to wind that is

Velocity of jet wrt wind= $V_{jw}$ = $V_{j}-V_{w}$

425 i = $V_{j}$ - 35 j

$V_{j}$ = 425 i + 35 j

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is,

∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

7 0

3 years ago

Which of the following is the best hypothesis

tekilochka [14]

Answer:

Melting and frezzing are physical changes

5 0

2 years ago

What is the purpose of oil used in a car's engine?

Hoochie [10]

B) to reduce friction

5 0

2 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

3 years ago

An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o

ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0

2 years ago