I need the answers to this please

the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

dem82 [27]

Answer:

Time = 5 seconds

Distance = 50 meters

Explanation:

Constantly Accelerated Motion

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

$v_f=v_o+a.t$

And the distance s is

$\displaystyle s=v_o.t+\frac{gt^2}{2}$

Given the object starts from rest, vo=0 and vf=20 m/s at $a=4\ m/s^2$ . We compute t

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}$

$\boxed{t=5\ sec}$

Now we compute s

$\displaystyle s=0+\frac{4\times 5^2}{2}$

$\boxed{s=50\ m}$

5 0

2 years ago

Michael tells Erica he weighs 104 pounds. Erica says she is 44 kilograms. If there are 2.2 pounds in a

Andrej [43]

Answer:

Michael is heavier.

Explanation:

If Michael is 104 pounds, that is equal to 47 kg.

Erica is 44 kg which is 97 pounds.

104 is greater than 97, and 47 is greater than 44.

6 0

3 years ago

Read 2 more answers

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

Read 2 more answers

Heating effect of current is not always useful for us. Support your answer with example

stellarik [79]

Answer:

is correct

Explanation:

in my think, first this due to ray emitted from the light those ray may be affect our skin or party of body.

4 0

3 years ago

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago