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vampirchik [111]

2 years ago

15

A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

nearest whole number.
What is the mass of the lawn mower?
kg
What is the normal force exerted on the lawn mower?
N

1 answer:

charle [14.2K]2 years ago

8 0

Answer:

are you in middle school

Explanation:

need more evendice

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A car accelerates at a constant rate of 3 m/s2 for 5 seconds. If it reaches a velocity of 27 m/s, what was its initial velocity?

Kaylis [27]

The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

CALCULATE INITIAL VELOCITY:

The initial velocity of the car can be calculated by using one of the equation of motion as follows:

V = u + at

Where;

V = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity (m/s²)
t = time (s)

According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:

u = v - at

u = 27 - 3(5)

u = 27 - 15

u = 12m/s.

Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

Learn more about motion at: brainly.com/question/974124

4 0

1 year ago

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

<em>Mass of the child, m = 16 kg</em>
<em>Length of the incline, L = 2 m</em>
<em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

$sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m$

The gravitational potential energy of the child at the top of the incline is calculated as;

$P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J$

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

1 year ago

A 2.00-kilogram object weighs 19.6 newtons on earth. if the acceleration due to gravity on mars is 3.71 meters per second2, what

Harrizon [31]

Recall that mass is the amount of matter present in a body. That means it's a property that is consistent regardless of the body's current location, gravity's pull on the body, etc.

Let's not confuse mass with weight (which is a force computed as Weight = mass x acceleration). Mass will remain constant and that means that whether the object is on Earth or on Mars, its mass remains the same. Thus, the object will still have 2.00 kg as mass on Mars.

Answers: 2.00 kilograms

4 0

3 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

2 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

3 years ago