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Misha Larkins [42]
2 years ago
10

if you already have the time which is 4s and the initial velocity which is 9.8m/s and the acceleration which is -9.8m/s/s. what

would the dispacement be?​
Physics
2 answers:
Vikentia [17]2 years ago
7 0

Answer:10

Explanation:

2+2=10

vichka [17]2 years ago
6 0

Answer:

10

Explanation:

2+8=10

6+4=10

5+5=10

7+3=10

9+1=10

3+2+4+1=10

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Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10
Naddika [18.5K]

This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V_D = ω_c × r_c  ----- let this be equ 1

where V_D is velocity of cylinder,  ω_c is the angular velocity of drum C and r_c is the radius of drum C

Now, Angular velocity of gear B is;

ω_B = ω_C

ω_B = V_D / r_c  -------- let this equ 2

so;

V_D / 0.1 m = 10V_D

Next, we determine the angular velocity of gear A;

from the diagram;

ω_A( 0.15 m ) = ω_B( 0.2 m )

from equation 2; ω_B = V_D / r_c

so

ω_A( 0.15 m ) = (V_D / r_c ) 0.2 m

substitutive in value of radius r_c (0.1 m)

ω_A( 0.15 m ) = (V_D / 0.1 m ) 0.2 m

ω_A( 0.15 ) = 0.2V_D / 0.1

ω_A =  2V_D  / 0.15

ω_A = 13.333V_D   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑U_{1-2 = T₂

0 + m_Dgh = \frac{1}{2}m_DV²_D + \frac{1}{2}I_Aω²_A + \frac{1}{2}I_Bω²_B

so

m_Dgh = \frac{1}{2}m_DV²_D + \frac{1}{2}(m_Ak_A²)(13.333V_D)² + \frac{1}{2}(m_Bk_B²)(10V_D)²

we given that; m_D = 50 kg, h = 2 m, m_A = 10 kg, k_A 125 mm = 0.125 m, m_B = 30 kg, k_B = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( \frac{1}{2} × 50 × V_D²) + \frac{1}{2}( 10 × (0.125)² )(13.333V_D)² + \frac{1}{2}( 30 × (0.15)²)(10V_D)²

981 = 25V_D² + 13.888V_D² + 33.75V_D²

981 = 72.638V_D²

V_D² = 981 / 72.638

V_D² = 13.5053

V_D = √13.5053

V_D = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s

7 0
3 years ago
The temperature of 2.0 g of helium is increased at constant volume by ΔT. What mass of oxygen can have its temperature increased
gavmur [86]

Answer:

m = 9.6 g

Explanation:

Thermal energy given to helium gas at constant volume is given as

Q = nC_v \Delta T

so here we have

C_v = \frac{3}{2}R

n = moles

n = \frac{2}{4} = 0.5

so we have

Q = \frac{3}{2}R(0.5)\Delta T

now we know that

for oxygen gas we have

C_v = \frac{5}{2}R

for same amount of heat we have

Q = nC_v \Delta T'

\frac{3}{2}R(0.5)\Delta T = \frac{m}{32} (\frac{5R}{2}) \Delta T

m = \frac{0.75 \times 32}{2.5}

m = 9.6 g

8 0
3 years ago
What is common to all fossil fuels?
natta225 [31]
All fossil fuels are formed deep underground. they are all nonrenewable energy
3 0
3 years ago
Read 2 more answers
A 50 kilogram woman wearing a seatbelt is traveling in a car that is moving with a velocity of +10 meters per second. In an emer
ra1l [238]

Answer:

the answer is A.) -1 * 10^3[N]

Explanation:

The solution consists of two steps, the first step is using the following kinematic equation:

v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]\\

The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].

Replacing:

0=10+a*(0.5)\\a=-20[m/s^2]

Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.

In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.

F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]

The minus sign means that the force is acting in the opposite direction of the movement.

7 0
3 years ago
A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. i
ollegr [7]

Magnetic field outside it due to long wire is given by

B = \frac{u_o i}{2 \pi r}

Magnetic field due to long wire inside wire at any point

B = \frac{u_o i r}{2 \pi R^2}

Now the ratio of two magnetic field is given by

\frac{B_{in}}{B_{out}} = \frac{r_1/R^2}{1/r_2}

\frac{0.285}{0.200} = \frac{4*10}{R^2}

1.425 = \frac{40}{R^2}

R = 5.3 mm

8 0
3 years ago
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