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kotykmax [81]
1 year ago
8

A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled

?
Physics
1 answer:
Leno4ka [110]1 year ago
8 0

Answer:

(a ) vf= 188.12m/s  : Final speed at 8.75 s

(b) d= 823.04 m   : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at            Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

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An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight
EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

v_{perpendicular} = v cos20

v_{parallel} = v sin20

so here we have

v_{parallel} = 16.2 sin20

v_{parallel} = 5.5 m/s

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0
2 years ago
2) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1.
Salsk061 [2.6K]
<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

<h3>pressure = 1.53 pascals</h3>

hope that helps and please tell me if i am wrong :)

8 0
1 year ago
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
Read 2 more answers
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
1 year ago
A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th
agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

a = 1 m/s^2

now for such acceleration we can use kinematics

d = \frac{1}{2}at^2

2 = \frac{1}{2}(1) t^2

t = 2 s

7 0
2 years ago
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