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3 years ago

What is the mass of an object that accelerates at 5 m/s2 when pushed with 100 N?

Physics

1 answer:

dangina [55]3 years ago

7 0

Answer:

the answer is 20g

Explanation:

Data:

acceleration:5m/s2

force:100N

mass:?

solution:

force: mass*acceleration

mass: force divided by acceleration

mass:100/5

mass:20g

or another way of solving it is

force: mass*acceleration

100:mass*5

100/5 : mass

20g:mass

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Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton

kodGreya [7K]

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

I = m(v-u)................. Equation 1

Where:

I = Impulse delivered to the baseball
m = mass of the baseball
v = Final velocity of the baseball
u = initial speed of the baseball

From the question,

⇒ Given:

m = 0.8 kg
u = 67 m/s
v = -44 m/s

⇒ Substitute these values into equation 1

I = 0.8(-44-67)
I = 0.8(-111)
I = -88.8
I ≈ -89 kgm/s

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

Learn more about impulse here: brainly.com/question/7973509

7 0

2 years ago

According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai

GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

$\Delta x=1.64\times 10^{-8}\ m$

So, the uncertainty in its position is $1.64\times 10^{-8}\ m$ . Hence, this is the required solution.

6 0

3 years ago

A fish tank is a cube of size L × L × L, where L = 1 m, filled with water. Find

Murljashka [212]

At the bottom of the tank :

P = ρgH

P = (1000 kg/m³)(10 m/s²)(1 m)

P = 10000 N/m²

F = P • A

F = (10000 N/m²)(1 m²)

F = 10000 N

At the side of the tank :

Pav = ½ρgH

Pav = ½(1000 kg/m³)(10 m/s²)(1 m)

Pav = 5000 N/m²

F = P • A

F = (5000 N/m²)(1 m²)

F = 5000 N

3 0

2 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

A new ski area has opened that emphasizes the extreme nature of the skiing possible on its slopes. Suppose an ad intones "Free-f

Allushta [10]

Answer:

48.6°

Explanation:

The forward force, F equals the component of the weight along the slope.

So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.

So a = gsinθ

Since we are given that a = 75%g = 0.75g,

0.75g = gsinθ

sinθ = 0.75

θ = sin⁻¹(0.75)

= 48.6°

5 0

3 years ago