Which shows the conversion of 2.09 × 10-4 meters to millimeters?

A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the

Brums [2.3K]

Answer:

$\boxed {\boxed {\sf 0.225 \ Newtons}}$

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= ma$

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

m= 0.75 kg
a= 0.3 m/s²

$F= 0.75 \ kg * 0.3 \ m/s^2$

Multiply.

$F =0.225 \ kg * m/s^2$

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

$F= 0.225 \ N$

<u>0.225 Newtons of force</u> are applied to the book.

5 0

2 years ago

A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def

Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0

3 years ago

A car has a mass of 550 kg. What is the weight of the car?

lozanna [386]

1212.54 is what 550 kgs is equal to the weight of the car

3 0

2 years ago

Read 2 more answers

Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.

alina1380 [7]

Answer:

The time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

Explanation:

Given the intensity of the energy is $1.36\ kW/m^2$

And the arriving energy is $3.55\times 10^9\ J$

Also, the area in which energy is being arriving is $4.25\ m^2$

Now, we will use relation between energy $(E)$ , intensity of energy $(p)$ , area $(A)$ and time $(T)$ .

Where energy is in Joule, intensity is in $kW/m^2$ , area is in $m^2$ and time is in seconds.

The equation is

$E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds$

So, the time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

4 0

3 years ago

A ramp 20 m long and 4 m high is used to lift a heavy box. A pulley system with 4 rope sections supporting the load is used to l

kobusy [5.1K]

The ramp has dimensions of 20 m length and 4 m height. The angle made by the ramp on the horizontal ground is
θ = sin⁻¹ (4/20)
After getting the angle, the force required to move the box can then be solved given that the mass of the box is known and also the coefficient of friction of the ramp

5 0

3 years ago