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2 years ago

an object with a mass of 7.5 kg accelerates 8.3 m/s2 when an unknown force is applied to it. what is the amount of the force?

Physics

2 answers:

3241004551 [841]2 years ago

6 0

Answer:

I believe the answer is 62.25 N

Explanation:

You just multiply 7.5 and 8.3 and you get 62.25

lana [24]2 years ago

5 0

Answer:

62.25N or 62.25 newtons

Explanation:

F=ma where F is the force in newtons, m is the mass in kilograms (kg), and a is the acceleration

m=7.5kg

a=8.3m/s^2

Therefore, F=ma=7.5*8.3=62.25N

The amount of force is 62.25N or 62.25 newtons

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Driving a motor vehicle requires many coordinated functions which are impacted by alcohol and other drugs

Anni [7]

I am assuming this is a true or false question, to which the answer would be True.

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2 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

2 years ago

Read 2 more answers

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

A box of mass 5.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s2 for 7.0 s. find the net work done

Aleks04 [339]

W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.

5 0

2 years ago

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A cart is pushed to the right with a force of 15 N while being pulled to the left with a force of 20 N. The net force on the car

9966 [12]

The net force of the cart when it is pushed to the right with a force of 15N.

<u>Explanation:</u>

To find the force of net, which is calculated by the formula.

The Net Force= Addition of the force applied on the respective direction.

The Net Force here is given by

The Net Force = 15-20 (A force towards the right and a force towards left, two opposite so subtraction).

Hence

Thus the Net Force = -5(The force towards left, so it gets a negative value).

5 0

2 years ago