A vehicle traveling at 10. m/s^2 accelerates at 1.5 m/s^2 for 10.s. What is the final velocity

4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable

icang [17]

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• m = 1 kg:

∑ F (parallel) = mg sin(45°) - T = ma … … … [1]

∑ F (perpendicular) = n - mg cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• m = 0.4 kg:

∑ F (vertical) = T - mg = ma … … … [2]

Adding equations [1] and [2] eliminates T, so that

((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a

(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a

==> a ≈ 2.15 m/s²

The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

8 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Does a weave with high-pitched have a very low frequency

Brilliant_brown [7]

No a wave with a high pitch has a very high frequency

8 0

3 years ago

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6. A 15 kg box is given an initial push so that it slides across the floor and comes to a stop.

Illusion [34]

Answer:

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

The displacement of the box can be determined using equation of motion as:

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.

4 0

4 years ago

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of it

liq [111]

Answer:

The acceleration of the laundry cart is 13.3 m/s²

Explanation:

Hi there!

According to Newton's second law, the sum of all forces acting on an object in a given direction is equal to the mass of the object times its acceleration in that direction:

∑F = m · a

Where:

∑F = net force acting on the cart.

m = mass of the cart.

a = acceleration.

Then, solving this equation for the acceleration:

∑F / m = a

60.0 N / 4.50 kg = a

a = 13.3 m/s²

The acceleration of the laundry cart is 13.3 m/s²

5 0

3 years ago

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