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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?

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 $_D$ gh = $\frac{1}{2}$ m $_D$ V² $_D$ + $\frac{1}{2}I_A$ ω² $_A$ + $\frac{1}{2}I_B$ ω² $_B$

so

m $_D$ gh = $\frac{1}{2}$ m $_D$ V² $_D$ + $\frac{1}{2}$ (m $_A$ k $_A$ ²)(13.333V $_D$ )² + $\frac{1}{2}$ (m $_B$ k $_B$ ²)(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 $_D$ ²) + $\frac{1}{2}$ ( 10 × (0.125)² )(13.333V $_D$ )² + $\frac{1}{2}$ ( 30 × (0.15)²)(10V $_D$ )²

981 = 25V $_D$ ² + 13.888V $_D$ ² + 33.75V $_D$ ²

981 = 72.638V $_D$ ²

V $_D$ ² = 981 / 72.638

V $_D$ ² = 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