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3 years ago

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Physics

2 answers:

V125BC [204]3 years ago

8 0

D. 128.25 because force=mass x acceleration

gtnhenbr [62]3 years ago

8 0

Answer:

D. 128.25 N

Explanation:

The force exerted on an object and the acceleration of the object are related by the following equation (Newton's second law):

$F=ma$

where

F is the force exerted on the object

m is the mass of the object

a is the acceleration

In this problem, we know:

- The mass of the object: m = 13.5 kg

- The acceleration: a = 9.5 m/s^2

Therefore, we can find the force required to give the object this acceleration:

$F=ma=(13.5 kg)(9.5 m/s^2)=128.25 N$

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A projectile is fored vertically upward with an initial velocity of 190 m/s. Find the maximum height of the projectile.

RSB [31]

Answer:

3683.67 m

Explanation:

Formula for maximum height of projectile is given by the equation;

h = u²/2g

Where u is initial velocity and g is acceleration due to gravity

We are given u = 190 m/s

Thus;

h = 190²/9.8

h = 36100/9.8

h = 3683.67 m

6 0

2 years ago

A diver rises quickly to the surface from a 5.0 m depth. If she did not exhale the gas from her lunds before rising, by what fac

Inessa [10]

Answer:

1.5 times

Explanation:

$h$ = depth of the diver initially = 5 m

$\rho$ = density of seawater = 1030 kg m⁻³

$P_{i}$ = Initial pressure at the depth

$P_{f}$ = final pressure after rising = 101325 Pa

Initial pressure at the depth is given as

$P_{i} = P_{f} + \rho gh\\P_{i} = 101325 + (1030) (9.8) (5)\\P_{i} = 151795 Pa$

$V_{i}$ = Initial volume at the depth

$V_{f}$ = Final volume after rising

Since the temperature remains constant, we have

$P_{f} V_{f} = P_{i} V_{i}\\(101325) V_{f} = (151795) V_{i}\\V_{f} = 1.5 V_{i}$

6 0

2 years ago

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te

Ber [7]

Answer:

Te = 23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

Te = 2462600 / 105322

Te = 23.4 °C

7 0

3 years ago

Read 2 more answers

What is TeO3 compound name

Stolb23 [73]

Tellurium trioxide

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4 0

2 years ago

Read 2 more answers

Which of the following maxims would apply to a person with a Type A personality?

lozanna [386]

The correct answer is C. Lost time is never found again

Explanation:

One of the most relevant personality theories is Type A and Type B personality theory that proposes two main types of personalities by grouping different personality traits. In the case of Type A personality, this refers to individuals that are very organized, impatient, and concerned with time and goals. On the opposite, Type B personality includes individuals that are more relaxed in different aspects.

According to this, one maxim that applies to Type A personality is "Lost time is never found again" because people with this personality are concerned about time and therefore, loss of time is considered highly negative by them. Also, due to their ambitions and concern with goals they want to avoid losing time as this is equivalent to work, money, goals, etc.

4 0

2 years ago