3 years ago

What is a push or pull that one object excerpts on another object?

Physics

2 answers:

Rasek [7]3 years ago

7 0

The answer would be force.

klasskru [66]3 years ago

5 0

Answer:

Force

Explanation:

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If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$

$\alpha = - 2.8 rad/s^2$

$t = 1.5 s$

now from kinematics we have

$\omega_f = \omega_i + \alpha t$

now from above all values we have

$\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)$

$\omega_f = -8.4 + (-4.2)$

$\omega_f = -12.6 rad/s$

so final angular speed is -12.6 rad/s

6 0

3 years ago

What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N

Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

= (45.0)×(7.50)

= 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

6 0

3 years ago

Identify characteristics of energy from the Sun. Check all that apply.

xxMikexx [17]

Answer:

First one, third one, and fourth one

6 0

2 years ago

Read 2 more answers

A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².

Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

$P=F/A$

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

$A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]$

Now replacing:

$P=100/0.0001\\P=1000000[Pa]$

3 0

3 years ago

A firecracker breaks up into three pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0

Mamont248 [21]

Answer:

$|P_3|=21.242m/s$

Explanation:

From the question we are told that:

Mass $m=200g=0.2$

Speed $v=82.0m/s$

Mass $M_2=300g=0.3kg$

Speed $V_2=45.0m/s$

Generally the equation for Magnitude of the Third piece is mathematically given by

$P_1+P_2+P_3=0$

$|P_3|=|P_1+P_2|$

Where

$P_1=m_1v_1$

$P_1=0.2*82$

$P_1=16.4kgm/s$

And

$P_2=0.3*45$

$P_2=13.5kgm/s$

Therefore

$|P_3|=\sqrt{16.4^2+13.5^2}$

$|P_3|=21.242m/s$

8 0

3 years ago