Ask question

Ask question

All categories

8 days ago

15

An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11

s. What will its velocity be after that acceleration?

1 answer:

serg [1.1K]8 days ago

7 0

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

You might be interested in

A 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s. If the

Keith_Richards [1034]

Answer:

No kinetic energy is lost as the collision is elastic.

Explanation:

Throughout an elastic collision, both momentum and kinetic energy remain conserved.

]} } ]} } ]} } } } ]}

6 0

8 days ago

Read 2 more answers

In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat

Yuliya22 [1153]

Answer:

-13.18°C

Explanation:

To solve this issue, we must examine the principles associated with the rate of thermal conduction.

This rate is defined by the equation

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = Amount of heat transferred

t = time

k = Thermal conductivity constant

A = Area of cross-section

$\Delta T =$ Temperature difference across the material

d = Material thickness

The scenario indicates a heat loss that is double the initial value, which means

$Q_2 = 2*Q_1$

Substituting values yields,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solving for $T_o$ ,

$T_o = -13.18$

Thus, when the heat lost per second is doubled, the temperature on the external surface of the window is -13.18°C.

3 0

11 days ago

At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee

Keith_Richards [1034]

Thrust is quantified as a reaction force, in accordance with Newton's third law. When a system accelerates or expels mass in one direction, this resulting mass generates a force of equal strength but in the opposite direction on that system. This relationship can be expressed mathematically as:

$T = v\frac{dm}{dt}$

Where:

v = velocity of the exhaust gases as perceived from the rocket.

$\frac{dm}{dt}$ = Change in mass over time

The provided data is as follows:

$v = 1500m/s$

$\frac{dm}{dt} = 100kg/s$

After substitution, we obtain:

$T = 1500*100$

$\therefore T = 1.5*10^5N$

7 0

6 days ago

A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk

kicyunya [1025]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

$r = 16m$

$m = 82kg$

$\mu_s = 0.63$

The maximum velocity can be determined using centripetal force,

$F_c = \frac{mv^2}{r}$

Should be equal to,

$\frac{mv^2}{r} = \mu_s mg$

$v = \sqrt{\mu_s gr}$

$v = \sqrt{(0.63)(9.8)(16)}$

$v = 9.93m/s$

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

3 0

13 hours ago

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

Keith_Richards [1034]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.

5 0

1 day ago