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3 years ago

Are we within earths Roche limit

1 answer:

3 0

Answer: Closer to the Roche limit, the body is deformed by tidal forces. Within the Roche limit, the mass' own gravity can no longer withstand the tidal forces, and the body disintegrates. hope this helps can u give me brainliest

Explanation:

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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

Which type of energy is commonly referred to as kinetic energy?

aksik [14]

Kinetic Energy is movement energy (most simplistic way I can put it) so its motion.

3 0

3 years ago

Read 2 more answers

A person is initially driving a car east down a straight road. the magnitude of the instantaneous acceleration is decreasing wit

Alexxandr [17]

By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,

4 0

3 years ago

Read 2 more answers

A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦

blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.

W= F*d *cosα Formula (1)

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force, inclined at θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7° (N)

Calculated of the FN

We apply the formula (2)

∑Fy = m*ay ay = 0

FN +Ty- W = 0

FN = W-Ty

FN = 89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*( 89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0

4 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

4 years ago