Can someone please help me ASAP please!

A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

padilas [110]

Explanation:

The weight of the car is equal to, $W_c=m\times g$ ...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, $F_N=mg\ cos\theta$

or

$F_N=mg\ cos(13)$ .............(2)

The horizontal component of the force is, $F_H=mg\ sin\theta$

Taking ratio of equation (1) and (2) as :

$\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}$

$\dfrac{F_N}{W_c}=cos(13)$

$\dfrac{F_N}{W_c}=0.97$

or

$\dfrac{F_N}{W_c}=\dfrac{97}{100}$

Hence, this is the required solution.

5 0

3 years ago

A person covers a distance of 320 miles in a travel time of 8 hours. What is the speed for this trip?

goldfiish [28.3K]

Take 320 and divide by 8
320/8 = 40 mi/hr or A

8 0

3 years ago

Read 2 more answers

A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the

Julli [10]

Answer:

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0

3 years ago

How many cubic centimeters of area does 762 g of distilled water occupy?

BartSMP [9]

1 kg of water = 1 L = 1 dm³
762 g of water = 762 cm³

8 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago