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RUDIKE [14]
3 years ago
6

A car driving down the road speeds up to pass bus. While passing changed velocity from 10 m/s to 35 m/s in 5 seconds. What is th

e acceleration of the car?
Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

The ans is 3

Explanation:

First write the given values and write the formula of acceleration than do velocity - final velocity by time taken is 10-35%5 so, therefore acceleration of car is 3

You might be interested in
Holding force constant, what will be the effect of increasing the Moment arm?
Murljashka [212]

Answer:

(b) Torque will increase.

Explanation:

Torque is given as the product of force and moment arm (radius).

τ = F x r

F = τ / r

where;

F is force

τ  is torque

r is radius (moment arm)

Keeping force constant, we will have the following;

τ ∝ r

This shows that torque is directly proportional moment arm (radius), thus increase in moment arm, will cause increase in torque.

For instance;

let the constant force = 5 N

let the initial moment arm, r = 2m

Torque, τ  = 5 N x 2m = 10 Nm

When the moment arm is increased to 4 m

Torque, τ  = 5 N x 4m = 20 Nm

Therefore, at a constant force, increasing in the Moment arm, will cause increase in torque.

Coorect option is "(b) Torque will increase."

5 0
3 years ago
6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces
Vesna [10]

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket V=15\ gal\approx 56.78\ L

time to fill it t=50\ s

Volume flow rate

\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s

The inner diameter of the hose D=1.5\ cm

diameter of the nozzle exit d=0.8\ cm

we can volume flow rate as

\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s

Mass flow rate

\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s

5 0
3 years ago
They occupy the space surrounding the nucleus of the atom, they have a -1 charge
kipiarov [429]

Answer:

Electrons.

Explanation:

Look it up.

4 0
3 years ago
During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on
IceJOKER [234]

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

3 0
3 years ago
a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box
tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , \theta=39^o.

Coefficient of kinetic friction , \mu=0.16.

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.

Hence , this is the required solution.

6 0
3 years ago
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