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dalvyx [7]
3 years ago
7

a 150 kg object takes a 1.5 minute travel to a 2500 straight path it begins you trip at traveling 120 miles an hour and it decel

erates to a velocity of 20 miles an hour what is it acceleration
Physics
1 answer:
crimeas [40]3 years ago
5 0
I believe it’s acceleration is -1.11 m/s^2
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leva [86]
Behaviorist approach to psychology
3 0
3 years ago
How do I find the cosine of theta.
NNADVOKAT [17]

The value of cos θ in the given figure is 0.98.

<h3>What is cosine of an angle?</h3>

The cosine of an angle is defined as the sine of the complementary angle.

The complementary angle equals the given angle subtracted from a right angle, 90.

cos θ = sin(90 - θ)

For example, if the angle is 30°, then its complement is 60°

cos 30 = sin(90 - 30)

cos 30 = sin 60

0.866 = 0.866

<h3>Cosine of an angle with respect to sides of a right triangle</h3>

cos θ = adjacent side / hypotenuse side

adjacent side of the given right triangle is calculated as follows;

adj² = 10² - 2²

adj² = 100 - 4

adj² = 96

adj = √96

adj = 9.8

cos θ = 9.8/10

cos θ = 0.98

Thus, the value of cos θ in the given figure is 0.98.

Learn more about cosine of angles here: brainly.com/question/23720007

#SPJ1

5 0
1 year ago
A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio
saveliy_v [14]

Answer: The coefficient of static friction is 3.85 and  The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

6 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =
Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

v = \omega R\\a = \alpha R

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft

One revolution is equal to the circumference of the drum. So, total number of revolutions is

x / (2\pi R) = 6/(\pi R)

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2

b) a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2

5 0
3 years ago
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