An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration

Answer:

The runner's deceleration is -23.33

Given:

Initial speed = 3.5

Final speed = 0

Time taken = 0.15 s

To determine:

Deceleration of the runner =?

Used Formula:

Using the first equation of motion,

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

Solution:

<pusing the="" first="" equation="" of="" motion="">

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

0 = 3.5 + a (0.15)

-3.5 = 0.15 (a)

a =

a = -23.33

The negative sign indicates that this represents deceleration.

Hence, the deceleration of the runner is -23.33

</pusing>

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

μk = 0.51, the coefficient of kinetic friction between the box and the roof

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (2.1)*a

7.039 - 9.86 = (2.1)*a

-2.821 = (2.1)*a

a=(-2.821)/(2.1)

a = -1.34 m/s²

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into Formula (2):

0² = v₀² + 2*(-1.34)*(5.4)

2*(1.34)*(5.4) = v₀²

v₀ = 3.8 m/s

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i + b j, where a and b represent constants to solve for.

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

By comparing the coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Thus,

B = 4.81452 i - 18.38908 j..... 4th quadrant

<pTherefore,

cos(Q) = 4.81452 / 12

Q = 66.346 degrees

360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.

Even if we lack details about the size of the pressure cooker or the altitude of its operation, we can reliably assess the force on the lid based on prior knowledge because, similar to boiling water, the pressure buildup inside the cooker increases in line with the rising temperature.