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3 years ago

A 6.0 kg object undergoes an acceleration of 2.0 m/s2 .

1 answer:

6 0

A. Fnet=ma
6*2=12N of force acting on the object in the direction it is accelerating

B. Fnet=ma
4*2=8N of force action on the object in the direction it is accelerating

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Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.

atroni [7]

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

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4 0

2 years ago

a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

kykrilka [37]

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

5 0

2 years ago

To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m

Lady bird [3.3K]

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

$P_{g}=\rho g h$

Where, $\rho$ =density

h = height

Put the value into the formula

$P_{g}=1040\times9.8\times4.94$

$P_{g}=50348.48\ Pa$

Pressure in atmospheres

$1\ atm =101.3\ kPa$

$P_{g}=\dfrac{50348.48}{101325}$

$P_{g}=0.4969\ atm$

Hence, The minimum gauge pressure is 0.4969 atm.

7 0

3 years ago

Read 2 more answers

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo

cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ( $a_c$ ) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ( $a_c$ ) is given by:

$a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s$

The velocity (v) is given by:

v = ωr; where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0

3 years ago

A mass hanged on a spring scale. what is the force exerted by gravity on 700g ?

Ipatiy [6.2K]

Answer:

6.86 N

Explanation:

Applying,

F = mg............... Equation 1

Where F = Force exerted by gravity on the mass, m = mass, g = acceleration due to gravity

Note: The Force exerted by gravity on the mass is thesame as the weight of the body.

From the question,

Given: m = 700 g = (700/1000) = 0.7 kg

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 9.8(0.7)

F = 6.86 N

6 0

2 years ago