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3 years ago

Describing Acceleration

Physics

2 answers:

vagabundo [1.1K]3 years ago

6 0

Answer: Negative acceleration occurs when an object slows down in the positive direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Explanation: Acceleration can be positive or negative. A negative acceleration is also known as deceleration. And deceleration occurs when an object is coming to rest. While acceleration simply means that an object is Speeding up.

Negative acceleration occurs when an object slows down in the positive direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Aleonysh [2.5K]3 years ago

5 0

Answer: A, C,D,F

A)Neg. acc. occurs when an object slows down in the pos. direct.

C)Neg. acc. occurs when an object slows down in the neg. direct.

D)Pos. acc. occurs when an object speeds up in the pos. direct.

F) Pos. acc. occurs when an object slows down in the neg. direct.

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A person drives a car around a circular road with a constant speed of 20 m/s. The

ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0

3 years ago

Read 2 more answers

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago

3. What is a manometer used to determine?

Gemiola [76]

Answer:

It's used to indicate pressure

3 0

3 years ago

Read 2 more answers

This chart shows characteristics of three different types of waves.

prohojiy [21]

Answer:

1. Primary or P waves are push and pull waves

2. Secondary, S or Shear Waves are also called transverse wave

3. L or surface waves reach the earth's surface after P and S waves

7 0

3 years ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago